Area equals Length times Width.
since the width of this equation is 5 units Longer then its Length
W = L + 5
To put this area in a equation we could say
L(L+5)=A
Simplify the equation, L^2 + 5L = A.
Unsure how to do this calculus, the book isn't explaining it well. Thanks
One way to capture the domain of integration is with the set
[tex]D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}[/tex]
Then we can write the double integral as the iterated integral
[tex]\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx[/tex]
Compute the integral with respect to [tex]y[/tex].
[tex]\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)[/tex]
Compute the remaining integral.
[tex]\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}[/tex]
We could also swap the order of integration variables by writing
[tex]D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}[/tex]
and
[tex]\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy[/tex]
and this would have led to the same result.
[tex]\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)[/tex]
[tex]\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)[/tex]
Item to ship 58 ,
items per case 6
Answer: 9 and 2/3 cases
Step-by-step explanation:
6 goes into 58 9 times to make 54 so makes 4 left over 4/6 simplified makes 2/3.
To ship 58 items with 6 items per case, you will need 9 complete cases and 1 additional case with the remaining 4 items, making a total of 10 cases.
We have,
To find the number of cases required to ship 58 items when there are 6 items per case, we can use division.
Number of cases = Total items / Items per case
Number of cases = 58 items / 6 items per case
Number of cases = 9 with a remainder of 4
When we divide 58 items by 6 items per case, we get 9 with a remainder of 4.
This means that 9 cases can be fully filled with 6 items each, and there will be 4 items left that won't fit in a complete case.
Therefore,
To ship 58 items with 6 items per case, you will need 9 complete cases and 1 additional case with the remaining 4 items, making a total of 10 cases.
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The complete question:
How many cases are required to ship 58 items if there are 6 items per case?
You are volunteering to help with the soccer team's Valentine's Day fundraiser. Each 16-ounce bag of nuts the team sold must include at least 60% chocolate-covered nuts inside.
However, instead of receiving a shipment of separated plain nuts and chocolate nuts, they delivered two large containers of mixed nuts. The first says it is 50/50 plain and chocolate covered. The second contains 80% chocolate-covered nuts.
The team is dismayed, but you come up with a solution. You suggest combining
ounces* of the 50/50 nuts with
ounces* of the 80% chocolate-covered nuts, to create the 60% mixture required for each bag.
*estimate
Then Bob, another student on the team says, "Wait, we promised at least 60%. So if we just do half and half, won't we be giving them at least 60%?"
Bob
correct.
We should combining 8 ounces of the 50/50 nuts with 8 ounces of the 80% chocolate-covered nuts, to create the 60% mixture required for each bag.
You are offering to assist with the Valentine's Day fundraiser for the soccer team, according to the query. At least 60% of the 16-ounce bags of nuts that the team sold had to be coated in chocolate.
However, they supplied two sizable containers of mixed nuts instead of a shipment that was divided into plain and chocolate nuts. First, it claims to be equally split between plain and chocolate-covered. The second one has almonds that are 80 percent chocolate-covered.
In order to create the 60% mixture required for each bag, we should combine 8 ounces of the 50/50 nuts with 8 ounces of the 80% chocolate-covered nuts.
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To make the 60 percent mixture needed for each bag, we should combine 8 ounces of the 50/50 nuts with 8 ounces of the 80 percent chocolate-covered nuts.
According to the question, you are willing to help with the Valentine's Day fundraiser for the soccer team. The team had to cover at least 60% of the 16-ounce bags of nuts they sold in chocolate.
Instead of a cargo that was split into plain and chocolate nuts, they instead provided two substantial containers of mixed nuts. First, it asserts that the basic and chocolate-covered portions are split equally. The second one has almonds that have been wrapped in 80 percent chocolate.
We need to combine 8 ounces of the 50/50 nuts with 8 ounces of the 80% chocolate-covered nuts.to make the 60 percent combination needed for each bag.
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WILL GIVE BRAINLIEST
HELP PLEASE
Select the correct answer.
The section of paper shown in the pattern below is of a circle. It will be wrapped around a cone. The wrapper will then be painted.
What is the amount of paint needed to paint 100 of these wrappers? Round your answer to the nearest hundred.
Answer:
C
Step-by-step explanation:
Area of a 1/4 sector of a circle = 1/4 * pi r^2
Givens
100 of these quarter sectors
pi = 3.14
r = 12
Solution
You are using area because you are going to paint each one of the sectors. You must cover every cm^2 with paint.
Area = 1/4 * 3.14 * r * r
Area = 1/4 * 3.14 * 12 * 12
Area = 113.25
But there are 100 of them. so the area is
100 * Area = 11325 cm^2
Rounded to the nearest 100 hundreds = 11300 cm^2
Answer:
The answer is 11,300 cm²
Step-by-step explanation:
I got it right on edmentum jope this helps your question.
Two sides of a triangle are 12cm and 8cm. Complete the inequality to show the possible lengths for the third side.
The inequalities that represents the third side of the triangle is c ≥ 4 or c ≤ 20
How to find the third side of a triangle?The triangle inequality theorem states that that the sum of any two sides of a triangle is greater than or equal to the third side.
Therefore, the two sides of the triangle are 12 cm and 8 cm.
A triangle with sides a, b and c follows the principle below;
a + b ≥ c.
Therefore,
let
c = third sides
12 + 8 ≥ c
20 ≥ c
c + 8 ≥ 12
c ≥ 4
c + 12 ≥ 8
Therefore, the third third should be greater than or equals to 4 or less than or equals to 20
c ≥ 4 or c ≤ 20
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If a b c d and m 2 is increased by 20 degrees how must m 3 be changed to keep the segments parallel?
The slope of another line is changed to 20 degree to keep the segments parallel.
According to the statement
We have given that the If ac is parallel to bd and m <5 is decreased by 20 degrees And we have to find that the how must m <2 be changed to keep the lines parallel.
So, For this purpose,
We know that the Slope or gradient of a line is a number that describes both the direction and the steepness of the line.
And here
The slopes of two lines will only same when these lines are parallel.
In other words, Only parallel lines have a same slope.
So, if slope of one line is decreased by 20 degree then we have to kept these line parallel. Then we have to decrease the slope of other line by 30 degree also.
So, The slope of another line is changed to 20 degree to keep the segments parallel.
Disclaimer: This question was incomplete. Please find the full content below.
Question:
If ac is parallel to bd and m <5 is decreased by 20 degrees how must m <2 be changed to keep the lines parallel?
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a chef uses 5 2/5 pounds of ground beef to make a hamburger and 2 1/4 pounds of ground beef to make meatloaf. the ground beef comes in 1/2 pound packages. how many packages of ground beef are needed to make hamburgers and meatloaf?
Answer:
16 packages
Step-by-step explanation:
5 2/5 + 2 1/4 Use the common denominator of 20
5 8/20 + 2 5/20
7 13/20
I know that I will need 14 packages for the 7 pounds. For the fraction part, 10/20 would be half, so I would need to buy a half pound for that, but I need 13/20 which would be more than a half pound, so I will need to buy 2 more pounds for the fraction part.
14 + 2 = 16
The scatter plot shows the relationship between the test scores of a group of students and the number of hours they exercise in a week:
On a grid, Label Hours of Exercise on x axis and Test Scores on y axis. The title of the graph is Test Scores and Exercise. The scale on the x axis shows the numbers from 0 to 10 at increments of 1, and the scale on the y axis shows numbers from 0 to 100 at increments of 10. Dots are made at the ordered pairs 1.1, 12 and 1, 30 and 2.1, 21 and 2.5, 42 and 3.5, 30 and 3.5, 45 and 4, 55 and 5, 45 and 5.5, 60 and 5.5, 70 and 6.5, 40 and 7, 50 and 7.5, 80 and 8, 75, and 8.5, 60 and 9, 75. The ordered pair 9, 15 is circled and labeled as T. All the other points are put in an oval and labeled as P.
Part A: What is the group of points labeled P called? What is the point labeled T called? Give a possible reason for the presence of point T. (5 points)
Part B: Describe the association between students' test scores and the number of hours they exercise. (5 points)
It should be noted that Group A is a cluster of points with similar property or relationship and Point B is an outlier. This is the point outside of the common relationship in the scatterplot
What is a scatterplot?It should be noted that scatter plots are the graphs that present the relationship between two variables in a data-set. It represents data points on a two-dimensional plane or on a Cartesian system.
In this case, the independent variable or attribute is plotted on the X-axis, while the dependent variable is plotted on the Y-axis.
The association between students' test scores and the number of hours they exercise is that there is a negative tendency. As number of hours spent on computer games increase, the test scores decrease.
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A printer prints an average of 2001/2 pages per hour. Approximately how many pages did it print in a 8-hour period
Answer:
Step-by-step explanation:
Comment
I can't be sure what you mean by 2001/2. I will interpret it the way it is written.
1 hour produces 2001/2
8 hours produces x
What you have is a proportion.
Solution
1/8 = 2001/2 // x Cross multiply
1*x = 2001/2 * 8 Combine
x = 8004
Answer
The printer will produce 8004 printed pages.
JOHN RICHARDS BORROWED $875 AGAINST HIS CHARGE ACCOUNT ON SEPTEMBER
12 AND REPAID THE LOAN ON OCTOBER 14 (32 DAYS LATER). IF THE INTEREST RATE
IS 0.04273% PER DAY, HOW MUCH INTEREST DID JOHN PAY ON THE LOAN AND WHAT
AMOUNT DID HE PAY THE BANK WHEN HE REPAID THE LOAN
Using simple interest, it is found that:
John paid $12 in interest for the loan.The total amount that he paid the bank was of $887.Simple InterestSimple interest is used when there is a single compounding per time period.
The amount of money after t units of time in is modeled by:
[tex]A(t) = A(0)(1 + rt)[/tex]
In which:
A(0) is the initial amount.r is the interest rate, as a decimal.For this problem, the parameters are given as follows:
A(0) = 875, r = 0.0004273, t = 32.
Hence the amount paid is given by:
A(32) = 875 x (1 + 0.0004273 x 32) = $887.
The amount of interest is given by:
I = 887 - 875 = $12.
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ASAPP help me with this question
Answer:
AD-DB
Step-by-step explanation:
Because ti joins altogether
Answer:
4th option
Step-by-step explanation:
given Δ ABD and Δ CBD are congruent then corresponding sides are congruent, that is
AB ≡ BC
Question
3x² + 25x - 18
Which of the following is a factor of the polynomial above?
Ox-9
O x + 3
O 3x - 2
O 3x + 1
The factors of polynomial [tex]3x^{2} +25x-18[/tex] is (3x-2)(x+9) that's why the correct option is (3x-2) which is option c.
Given a polynomial [tex]3x^{2} +25x-18[/tex].
We are required to find the factors of polynomial given as [tex]3x^{2} +25x-18[/tex].
Polynomial is a combination of algebraic terms which is formed by using algebraic operations.
Factors are those numbers which when divided gives the number whose factors they are.
[tex]3x^{2} +25x-18[/tex]
To find the factors we need to break the middle term of the polynomial so that the broken parts when multiplied gives the product of both the first term and last term.
[tex]3x^{2}[/tex]+27x-2x-18
3x(x+9)-2(x+9)
(3x-2)(x+9)
Hence the factors of polynomial [tex]3x^{2} +25x-18[/tex] is (3x-2)(x+9) that's why the correct option is (3x-2) which is option c.
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someone please help me in this question i need step by step sol with explaination
Converting the portions to decimal, the grade that has the greatest portion of students is: 3rd grade.
How to Compare Percentage, Decimals, and Fractions?The bigger the digit that is close to the decimal point, the bigger the number given.
Also, percentage can be converted to decimal by dividing the given percent by 100. Percent means over 100.
Fractions can also be converted to decimals by simply dividing the numerator by the denominator.
To compare the portion of students interested in playing an instrument in each grade, let all the portions be in decimals.
Converting 25/36 to decimal, we have: 0.694 (3rd grade)
Converting 61.24% to decimal, we have: 61.24/100 = 0.6124 (4th grade)
From the rest of the portions given, 0.694 is greater than the others.
Therefore, the grade that has the greatest portion of students is: 3rd grade.
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What is the perimeter of angle DEF to the nearest tenth of a unit?
helppppp
tysm
===========================================================
Explanation:
Let's use the distance formula to find the distance from D to E
[tex]D = (x_1,y_1) = (-7,2) \text{ and } E = (x_2, y_2) = (3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-7-3)^2 + (2-4)^2}\\\\d = \sqrt{(-10)^2 + (-2)^2}\\\\d = \sqrt{100 + 4}\\\\d = \sqrt{104}\\\\d = \sqrt{4*26}\\\\d = \sqrt{4}*\sqrt{26}\\\\d = 2\sqrt{26}\\\\d \approx 10.198\\\\[/tex]
Note: uppercase D refers to the point, while lowercase d is the distance from D to E.
The length of segment DE is roughly 10.198 units long.
----------------
Repeat for the distance from E to F.
[tex]E = (x_1,y_1) = (3,4) \text{ and } F = (x_2, y_2) = (5,-7)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-5)^2 + (4-(-7))^2}\\\\d = \sqrt{(3-5)^2 + (4+7)^2}\\\\d = \sqrt{(-2)^2 + (11)^2}\\\\d = \sqrt{4 + 121}\\\\d = \sqrt{125}\\\\d = \sqrt{25*5}\\\\d = \sqrt{25}*\sqrt{5}\\\\d = 5\sqrt{5}\\\\d \approx 11.1803\\\\[/tex]
Segment EF is roughly 11.1803 units long.
----------------
Repeat for the distance from F to D.
[tex]F = (x_1,y_1) = (5,-7) \text{ and } D = (x_2, y_2) = (-7,2)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(5-(-7))^2 + (-7-2)^2}\\\\d = \sqrt{(5+7)^2 + (-7-2)^2}\\\\d = \sqrt{(12)^2 + (-9)^2}\\\\d = \sqrt{144 + 81}\\\\d = \sqrt{225}\\\\d = 15\\\\[/tex]
Unlike the others, this result is exact.
----------------
Add up the three segment lengths to get the perimeter
DE + EF + FD
10.198 + 11.1803 + 15
36.3783
The perimeter is approximately 36.3783 units which rounds to 36.4
The answer has been confirmed with GeoGebra.
Find the magnitude and direction of each resultant for the given vectors. Round each side to the nearest tenth and round each angle to the nearest degree.
w = < 5, 6 >, x = < -1, 14 >
The magnitude of the resultant vectors is 20.4 units and the direction of the vectors is 79⁰.
Magnitude of the resultant vector
Sum of vectors in x direction, ∑X = 5 - 1 = 4
Sum of the vectors in y direction, ∑Y = 6 + 14 = 20
Resultant vector = √(∑X² + ∑Y²)
Resultant vector = √(4² + 20²) = 20.4 units
Direction of the vectorθ = arc tan (ΣY/ΣX)
θ = arc tan (20/4)
θ = arc tan (5)
θ = 78.7⁰
θ ≈ 79⁰
Thus, the magnitude of the resultant vectors is 20.4 units and the direction of the vectors is 79⁰.
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Find the surface area of the composite figur
2 cm
5 cm
4 cm
3 cm
5 cm
SA =
4 cm
3 cm
5 cm
[?] cm²
By definition of surface area and the area formulae for squares and rectangles, the surface area of the composite figure is equal to 166 square centimeters.
What is the surface area of a composite figure formed by two right prisms?
According to the image, we have a composite figure formed by two right prisms. The surface area of this figure is the sum of the areas of its faces, represented by squares and rectangles:
A = 2 · (4 cm) · (5 cm) + 2 · (2 cm) · (4 cm) + (2 cm) · (5 cm) + (3 cm) · (5 cm) + (5 cm)² + 4 · (3 cm) · (5 cm)
A = 166 cm²
By definition of surface area and the area formulae for squares and rectangles, the surface area of the composite figure is equal to 166 square centimeters.
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P = √mx_ £x MX t make x the subject
[tex]p = \sqrt{ \frac{mx}{t} } - t {}^{2}x \: \: make \: x \: the \: subject [/tex]
So, making x subject of the formula, x = [m - 2pt³ ±√(m² - 4pt²m)]/{2t⁵}
How to make x subject of the formula?Since p = √(mx/t) - t²x
So, p + t²x = √(mx/t)
Squaring both sides, we have
(p + t²x)² = [√(mx/t)]²
p² + 2pt²x + t⁴x² = mx/t
Multiplying through by t,we have
(p² + 2pt²x + t⁴x²)t = mx/t × t
p²t + 2pt³x + t⁵x² = mx
p²t + 2pt³x + t⁵x² - mx = 0
t⁵x² + 2pt³x - mx + p²t = 0
t⁵x² + (2pt³ - m)x + p²t = 0
Using the quadratic formula, we find x.
[tex]x = \frac{-b +/-\sqrt{b^{2} - 4ac} }{2a}[/tex]
where
a = t⁵, b = (2pt³ - m) and c = p²tSubstituting the values of the variables into the equation, we have
[tex]x = \frac{-(2pt^{3} - m) +/-\sqrt{(2pt^{3} - m)^{2} - 4(t^{5})(p^{2}t) } }{2t^{5} }\\= \frac{-(2pt^{3} - m) +/-\sqrt{4p^{2} t^{6} - 4pt^{2}m + m^{2} - 4p^{2}t^{6} } }{2t^{5}}\\= \frac{-(2pt^{3} - m) +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}\\= \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}[/tex]
So, making x subject of the formula, [tex]x = \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}[/tex]
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J(1)²,
x ≤ 0
x > 0
2²,
Complete the table of values for function f, and then plot the ordered pairs on the graph.
-2
-1
0
1
2
f(x)
=
X
f(x)
Drawing Tools
Select
Point
Open Point
>
O
Click on a tool to begin drawing.
-10
-8
-6
-4
-2
10
8
6
4-
2-
-2
पं
CC
f(x)
Delete
2
4
Undo
6
3
+00
8
Answer:
1/2 - 4, 1/2 2, 2 1/4,
Step-by-step explanation:
1/2, - 4 1/2 2,2,1/4
The table when completed looks like this:
x -2 -1 0 1 2
f(x) 4 2 1 2 4.
We plot these points on the graph: (-2, 4), (-1, 2), (0, 1), (1, 2), (2, 4).
What is a function?A function exists as a relation between a dependent variable (f(x)) and an expression of an independent variable (x), utilized to define the value of a dependent variable from a provided value of an independent variable.
We have been given a function,
[tex]$f(x)=\left \{ {{(1/2)^{x},x\leq 0 } \atop {2^x > 0}} \right.[/tex]
We have to estimate the value of f(x) when x = {-2, -1, 0, 1, 2}
If the values of x ≤ 0, take f(x) = (1/2)ˣ then
f(-2) = (1/2)⁻² = 2² = 4
f(-1) = (1/2)⁻¹ = 2¹ = 2
f(0) = (1/2)⁰ = 1
If the values of x > 0, take f(x) = 2ˣ then
f(1) = 2¹ = 2
f(2) = 2² = 4
The table exists as follows
x -2 -1 0 1 2
f(x) 4 2 1 2 4.
We plot these points on the graph: (-2, 4), (-1, 2), (0, 1), (1, 2), (2, 4).
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The controller of MingWare Ceramics Inc. wishes to prepare a cost of goods sold budget for September. The controller assembled the following information for constructing the cost of goods sold budget: Direct materials: Enamel Paint Porcelain Total Total direct materials purchases budgeted for September $35,870 $7,530 $139,890 $183,290 Estimated inventory, September 1 1,730 4,150 6,920 12,800 Desired inventory, September 30 2,980 2,710 7,150 12,840 Direct labor cost: Kiln Department Decorating Department Total Total direct labor cost budgeted for September $51,550 $149,500 $201,050 Finished goods inventories: Dish Bowl Figurine Total Estimated inventory, September 1 $5,810 $3,310 $2,730 $11,850 Desired inventory, September 30 3,720 4,590 4,360 12,670 Work in process inventories: Estimated inventory, September 1 $3,540 Desired inventory, September 30 1,980 Budgeted factory overhead costs for September: Indirect factory wages $80,400 Depreciation of plant and equipment 10,550 Power and light 5,110 Indirect materials 3,390 Total 99,450 Use the preceding information to prepare a cost of goods sold budget for September. For those boxes in which you must enter subtracted or negative numbers use a minus sign. MingWare Ceramics Inc. Cost of Goods Sold Budget For the Month Ending September 30 Finished goods inventory, September 1 $Finished goods inventory, September 1 Direct labor $Direct labor Direct materials: $- Select - - Select - $- Select - - Select - $- Select - Direct labor fill in the blank 15 - Select - - Select - $- Select - Work in process inventory, September 30 fill in the blank 22 - Select - $- Select - - Select - $- Select - 4,100
The preparation of the Cost of Goods Sold Budget for MingWare Ceramics Inc. in September is as follows:
Cost of Goods Sold Budget:Beginning inventory of finished goods $11,850
Cost of goods manufactured 485,310
Ending inventory of finished goods (12,670)
Cost of Goods Sold $484,490
What is the cost of goods sold budget?The cost of goods sold budget sums the costs of the beginning inventory of finished goods with the cost of goods manufactured and subtracts the ending inventory of finished goods.
The cost of goods sold budget is based on the cost of goods manufactured budget.
Data and Calculations:Direct materials: Enamel Paint Porcelain Total
Total direct materials purchases
budgeted for September $35,870 $7,530 $139,890 $183,290
Estimated inventory, September 1 1,730 4,150 6,920 12,800 Desired inventory, September 30 (2,980) (2,710) (7,150) (12,840)
Materials used in production $183,250
Direct labor cost: Kiln Department Decorating Department Total
Total direct labor cost
budgeted for September $51,550 $149,500 $201,050
Finished goods inventories: Dish Bowl Figurine Total
Estimated inventory, September 1 $5,810 $3,310 $2,730 $11,850
Desired inventory, September 30 3,720 4,590 4,360 12,670
Cost of goods manufactured budget:Work in process inventories:
Estimated inventory, September 1 $3,540
Direct materials used in production $183,250
Direct labor costs $201,050
Total factory overhead costs $99,450
Desired inventory, September 30 (1,980)
Cost of goods manufactured $485,310
Budgeted factory overhead costs for September:Indirect factory wages $80,400
Depreciation of plant and equipment 10,550
Power and light 5,110
Indirect materials 3,390
Total 99,450
Thus, to prepare the cost of goods sold, as above, MingWare Ceramics, requires to determine the cost of goods manufactured for September.
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Can someone please help me with this?
Answer: Option (2)
Step-by-step explanation:
When x=10,
The median-median line gives [tex]y=1.4(10)+2.6=16.6[/tex].The least-squares regression line gives [tex]y=0.9(10)+4.2=13.2[/tex].13.2 is closer to 14 than 16.6, so the least-squares regression line is a better prediction.
Volume oblate spheroid
The volume of the oblate spheroid is approximately equal to 75.398 cubic feet.
What is the volume of an oblate spheroid?
In this problem we have the shape of an ellipse centered at origin, whose vertex form is shown below:
x² / a² + y² / b² = 1 (1)
Where a, b are the lengths of the semiaxes, in feet.
An oblate spheroid is generated by revolving half of the ellipse about the y-axis. Oblate spheroids are a kind of ellipse:
x² / a² + y² / b² + z² / a² = 1 (2)
Where a, b, c are the lengths of the semiaxes, in feet.
And the volume of the oblate spheroid is:
V = (4 / 3) · π · a² · b (3)
If we know that a = 3 ft, b = 2 ft, then the volume of the oblate spheroid is:
V = (4 / 3) · π · (3 ft)² · (2 ft)
V ≈ 75.398 ft³
The volume of the oblate spheroid is approximately equal to 75.398 cubic feet.
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The average of six positive integers starting with a is equal to b. What is the average of five consecutive integers ending with b?
a)a+1
b)a-1
C)a+2
d)a+3
e)a+4
The average of the five consecutive numbers ending with b in discuss when expressed in terms of a is; Choice D; a+3.
What is the average of five consecutive integers ending with b?First, since it was given in the task content that the average of six positive consecutive odd integers starting with a is equal to b, it therefore follows that;
(a+a+2+a+4+a+6+a+8+a+10)/6 = b
6b=6a+30
b=a+5
Also, let the average of the consecutive intergers ending with b be denoted by; x.
(b+b-1+b-2+b-3+b-4)/5 = x
=(5b-10)/5
=b–2
The average, x=b – 2 (where b = a-5)
Ultimately, the value of the required average is; = a+5-2 = a+3.
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When a force of 60 N acts on a certain object, the acceleration of the object is 6/ms2. if the acceleration of the object becomes 7/ms2, what is the force?
When the acceleration is 7 m/s², it means that the force will be 70 N.
How to find the force on an object?From newton's first law of motion, we know that;
F = ma
where;
m is mass
a is acceleration
We are given;
F = 60 N
a = 6 m/s²
Thus;
m = F/a = 60/6
m = 10 kg
When acceleration is 7 m/s², we have;
F = 10 * 7
m = 70 N
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Will mark brainliest
a. The area on [0, 10] is that of a trapezoid with bases 5 and 15 and height 10, so
[tex]\displaystyle \int_0^{10} f(x) \, dx = \frac{5+15}2\cdot10 = \boxed{100}[/tex]
b. By linearity of the definite integral, we have
[tex]\displaystyle \int_0^{25} f(x)\,dx = \int_0^{10} f(x)\,dx + \int_{10}^{25} f(x)\,dx[/tex]
and the area on [10, 25] is another trapezoid with bases 15 and 7.5 and height 15, so that
[tex]\displaystyle \int_{10}^{25} f(x)\,dx = \frac{15+7.5}2\cdot15 = 168.75[/tex]
Then the total area on [0, 25] is
[tex]\displaystyle \int_0^{25} f(x)\,dx = \boxed{268.75}[/tex]
c. The area on [25, 35] is that of a triangle with base 10 and height 15. However, [tex]f(x)<0[/tex] on this interval, so we multiply this area by -1 to get
[tex]\displaystyle \int_{25}^{35} f(x)\,dx = -\frac{10\cdot15}2 = \boxed{-75}[/tex]
d. The area on [15, 25] is the same as the area on [25, 35] because it's another triangle with the same dimensions. But the area on [15, 25] lies above the horizontal axis, so
[tex]\displaystyle \int_{15}^{25} f(x)\,dx = \int_{15}^{25} f(x)\,dx + \int_{25}^{35} f(x)\,dx = \boxed{0}[/tex]
e. The plot of [tex]|f(x)|[/tex] lies above the horizontal axis. We know the area on [15, 25] is the same as the area on [25, 35], but now both areas are positive, so
[tex]\displaystyle \int_{15}^{35} |f(x)| \, dx = \int_{15}^{25} f(x)\,dx - \int_{25}^{35} f(x)\,dx = 2 \int_{15}^{25} f(x)\,dx = \boxed{150}[/tex]
f. Changing the order of the limits in the integral swaps the sign of the overall integral, so
[tex]\displaystyle \int_{10}^0 f(x)\,dx = -\int_0^{10} f(x)\,dx = \boxed{-100}[/tex]
ASAP help me with this question
Answer:
SAS and SSA
Kindly award branliest
[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]
Congruency refers to the criteria of making two identical shapes that can hide each other properly when overlapped.
The criterias that guarantee congruence are :
SSSSASASAAAS[tex] \qquad \large \sf {Conclusion} : [/tex]
Correct options are : b, c, d, eQuestions are in the picture
The value of x which S(x) is a global minimum is x = 1/2
Find a formula for S(x)From the question, we have:
x is a positive numberthe sum of its reciprocal and four times the product of x is the smallest possibleThis means that:
S(x) = 1/x + 4x^2
The domain of xFrom the question, we understand that x is a positive number.
This means that the domain of x is x > 0
As a notation, we have (0, ∞)
The value of x which S(x) is a global minimumRecall that:
S(x) = 1/x + 4x^2
Differentiate the function
S'(x) = -1/x^2 + 8x
Set to 0
-1/x^2 + 8x = 0
Multiply through by x^2
-1 + 8x^3 = 0
Add 1 to both sides
8x^3 = 1
Divide by 8
x^3 = 1/8
Take the cube root of both sides
x = 1/2
To prove the point is a global minimum, we have:
S'(x) = -1/x^2 + 8x
Determine the second derivative
S''(x) = 2/x^3 + 8
Set x = 1/2
S''(x) = 2/(1/2)^3 + 8
Evaluate the exponent
S''(x) = 2/1/8 + 8
Evaluate the quotient
S''(x) = 16 + 8
Evaluate the sum
S''(x) = 24
Because S'' is positive, then the single critical point is a global minimum
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3x^2 + 12x - 96 help please
Answer:
fully factored form = 3(x-4)(x+8)
Step-by-step explanation:
first you factor 3 out
3(x^2+4x-32)
then factor it
3(x-4)(x+8)
Answer:
3(x-4)(x+8) This is the answer :)
Step-by-step explanation:
3x^2+12x - 96
3(x^2+4x-32)
3(x^2+4x--32)
3(x^2+8x-4x-32)
3(x^2+8x-4x-32)
3(x(x+8)-4(x+8))
3(x(x+8)-4(x+8))
3(x-4)(x+8)
3(x=4)(x+8)
6
Select the correct answer.
A company polled a large group of people to assess the company's reputation in the surrounding communities. To save time, the survey was
mailed to the addresses of stockholders and former employees that the company already had on file and that were in the surrounding areas.
Which of the following statements is correct?
Answer:
a company powder large group of people to assess the company's reputation in the surrounding
On parallelogram ABC∧
mΔA=42 mΔB=___ mΔC___ mΔD=___
The measures of the remaining angles of the parallelogram are m ∠ B = 138°, m ∠ C = 42° and m ∠ D = 138°
How to determine the measure of missing angles in a parallelogram by geometric theorems
In this problem we know the measure of an angle and we should find the values of the three remaining angles. According to Euclidean geometry, the sum of internal angles in a parallelogram equals 360° and there exists the following property between each pair of angles:
m ∠ A = m ∠ Cm ∠ B = m ∠ Dm ∠ A + m ∠ B = m ∠ B + m ∠ C = m ∠ C + m ∠ D = m ∠ A + m ∠ D = 180°Then, the measures of the missing angles are, respectively:
m ∠ B = 180° - m ∠ A
m ∠ B = 180° - 42°
m ∠ B = 138°
m ∠ D = 138°
m ∠ C = 42°
Remark
The statement is poorly formatted, presents typing mistakes and image is missing, correct form is shown below:
On parallelogram ABCD, m ∠ A = 42, m ∠ B = ____, m ∠ C = _____, m ∠ D = _____
A representation of the parallelogram is shown in the image attached below.
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Merina is scheduled to make two loan payments to Bradford in the amount of $1,000 each, two months and nine months from now. Merina doesn't think she can make those payments and offers Bradford an alternative plan where she will pay two equal amounts seven months from now followed by another equal payment seven months later. Bradford determines that 8.5% is a fair interest rate. Using the second unknown equal payment as your focal date, what are the equal payments?
well, we're assuming all along that Merina owes Bradford $2000, because in the 1st scenario, she was going to pay twice $1000.
on the 2nd scenario, she'll be paying the same $2000 but split 7 months from now and then 7 months later, same 2000 bucks, at which point Bradford applied 8.5% interest.
using those assumptions, since the wording is not quite clear, we can say that Merina is simply paying 2000 bucks plus the 8.5%
[tex]\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{8.5\% of 2000}}{\left( \cfrac{8.5}{100} \right)2000}\implies 170 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\stackrel{principal}{2000}~~ + ~~\stackrel{interest}{170}}{2}\implies \stackrel{\textit{two equal payments of}}{1085}[/tex]