[tex]s = c(1 + in) \\ s = 4000(1 + 0.03 \times 1) [/tex]
[tex]s = 4120.00[/tex]
Option: Asomeone 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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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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how much money would u get from $521 a second in 19 minutes
We get $593,940
Answer:
Solution Given:
1 second = $ 521
19 minutes=19*60=1140 seconds
1140 seconds =$521*1140 =$593,940
Answer:
$593,940
Step-by-step explanation:
$521 for every second
First, you need to multiply 60 with 19, since there are 60 seconds in a minute.
We get: 60 × 19 = 1140
Now multiply $521 with 1140
We get: 521 × 19 = 593,940
Final result: $593,940
Hope this helps :)
Coach Kirby is forming dodgeball teams from students in her physical education class.
She wants teams that each have 6 students. There are fewer than 48 students in the class,
so Coach Kirby can't form as many full teams as she wants.
1)) Let x represent how many full teams Coach Kirby can form. Which inequality describes
the problem?
6x < 48
6x > 48
Solve the inequality. Then, complete the sentence to describe the solution.
1) Coach Kirby can form fewer than 8|
full teams.
Please hurry
Answer:
6x<48
Step-by-step explanation:
ik its this one because theres 48 players and 6 teams both are even numbers so if you did 48÷6x you would get 8 but theres fewer than 48 kids
A sequence starts 1, -1 . Give a different rule the sequence could follow and the next 3 terms.
7/3 4/2 as improper fraction
Answer:
If the question asked mixed fraction:
7/3 = 2 1/3
4/2 = 2
Step-by-step explanation:
I converted it to mixed fraction because it is already improper fraction.
It takes Natalie's mother 20 minutes to drive to work. She works 6 miles away.
If she drives the same speed to pick up Natalie at a friend's house that is 14 miles
away, about how long will the drive take?
A 8.7 minutes
B 2.4 minutes
C 45.7 minutes
D 705 minutes
Answer:
B. 45.7
Step-by-step explanation:
20/6=3.33
3.33*14=46.62
46.62 is closest to 45.7
Therefore, 45.7 is the best answer.
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.
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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SOMEBODY PLEASE HELP ASAP
Answer:
40
Step-by-step explanation:
[tex]\frac{RW}{24}=\frac{30}{18} \\ \\ RW=40[/tex]
Mrs. Oliver drew a box plot to represent her students’ scores on a mid-term test.
Answer:
About 1/4 scored higher and 3/4 scored lower
Step-by-step explanation:
84 is on the right side line
Each line represents 25 % of the score
It is 75 percent ( 3 lines) of the way
She scored higher than 75% of the students and lower than 25% of the students
About 1/4 scored higher and 3/4 scored lower
e) 1 cubic centimetres = 1÷1000000 cubic meter. Rewrite in scientific notation
[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
Here we go ~
[tex]\qquad \sf \dashrightarrow \: 1 \: \: cm {}^{3} = \cfrac{1}{1000000} \: \: m {}^{3} [/tex]
[tex]\qquad \sf \dashrightarrow \: 1 \: \: cm {}^{3} = \cfrac{1}{10 {}^{6} } \: \: m {}^{3} [/tex]
[tex]\qquad \sf \dashrightarrow \: 1 \: \: cm {}^{3} = {}{10 {}^{ - 6} } \: \: m {}^{3} [/tex]
What number is next 3,8,35,48,99
Answer: 120
Step-by-step explanation: The pattern is a very weird pattern, but still a pattern.
The pattern starts with number 2 to the power of 2, and then subtract 1. This is 3.
Then we keep using this same pattern, keeping the exponent the same but increasing the numbers. 3^2 is 9, and subtracting 1 gives 8.
Then the pattern skips two numbers 4 and 5. So for every 2 numbers that is squared and subtracted 1, 2 other numbers are not apart of the pattern.
So the next is 6^2, which is 36-1 which is 35.
Then 7^2 is 49 and subtracting 1 is 48.
Then, ignoring the next 2 numbers because of the pattern, we have 10^2 which is 100, and subtracting 1 is 99.
Now the next number is 11^2 which is 121, and subtracting 1 is 120.
Hope this helped! If it did, please mark as Brainliest.
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]
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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1.
Find a polynomial f(x) of degree 3 that has the following zeros.
4, 0, -7
Leave your answer in factored form.
Answer:
x³+3x²-28x
Step-by-step explanation:
(x-4)(x-0)(x+7)
(x-4)(x)(x+7)
(x-4)(x+7)=(x²+3x-28)
(x²+3x-28)(x)=x³+3x²-28x
Questions 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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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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A person needs to fill 15 water jugs with a hose. Filling the first 3 jugs has taken 4 minutes. How long to finish filling the remaining jugs?
Answer:
20 minutes
Step-by-step explanation:
Rate x Time = jobs
The rate is 3 jugs/ 4 minutes.
The number of jobs is 15
Use T for time
3/4T = 15 Multiply both sides by 4/3 to get the variable T by itself
(4/3) 3/4 T = 15(4/3)
T = 20 minutes
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
At the beginning of 2000 Randall's house was worth 234 thousand dollars and Damien's house was worth 110 thousand dollars. At the beginning of 2003, Randall's house was worth 176 thousand dollars and Damien's house was worth 154 thousand dollars. Assume that the values of both houses vary at an exponential rate.
Write a function [tex]f[/tex] ,that determines the value of Randall's house (in thousands of dollars) in terms of the number of years [tex]t[/tex] since the beginning of 2000.
Write a function [tex]g[/tex] that determines the value of Damien's house (in thousands of dollars) in terms of the number of years [tex]t[/tex] since the beginning of 2000.
How many years after the beginning of 2000 will Randall's and Damien's house have the same value?
Using linear functions, we have that:
Randall's value is: f(t) = -19t + 234.Damien's value is: g(t) = 14.67t + 110They will have the same value in 3.68 years since the start of 2000.A linear function is modeled by:
y = mx + b
In which:
m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.We consider the initial values as the y-intercepts. Randall's house decayed 58 thousand dollars in 3 years, hence the slope is:
m = -58/3 = -19.
Hence the function for the value, in thousands of dollars, is:
f(t) = -19t + 234.
Damien's initial value was of 110, and it increased 44 thousand in 3 years, hence the slope is:
m = 44/3 = 14.67
Hence the function for the value of Damien's house, in thousands of dollars, in t years after 2003, is:
g(t) = 14.67t + 110
They will have the same value when:
f(t) = g(t)
-19t + 234 = 14.67t + 110
33.67t = 124
t = 124/33.67
t = 3.68
They will have the same value in 3.68 years since the start of 2000.
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A rectangular table is six times as long as it is wide. If the area is 150 ft2, find the length and the width of the table.
The width of the table is
The length of the table is?
Answer:
Length = 30ft, Width = 5ft
Step-by-step explanation:
Let x be the width.
Area of Rectangle = Length * Width
Given from the information in the question,
Length = 6x ft
Width = x ft
Substitute the values into the formula:
6x * x = 150
[tex]6x^{2}[/tex] = 150
[tex]x^{2} =\frac{150}{6}[/tex]
[tex]x^{2} =25[/tex]
[tex]x=\sqrt{25}[/tex]
x = 5 ft.
Therefore,
Length = 6 * 5 = 30ft
Width = 5 ft.
The following formula models the number of automobiles, A,in thousands, that Washington residents purchased
x years after 1980.
A=2x2+24x+300
A) Write an equation that you can use to determine the first year in which residents purchased approximately 566 thousand cars.
B) Now solve your equation and specify the first year (for example, enter 2019 and not 39) in which residents purchased approximately 566 thousand cars.
Considering the given quadratic equation for the number of automobiles, we have that:
A) The equation to be solved is: x² + 12x - 133 = 0
B) The first year was of 1987.
What is a quadratic function?A quadratic function is given according to the following rule:
[tex]y = ax^2 + bx + c[/tex]
The solutions are:
[tex]x_1 = \frac{-b + \sqrt{\Delta}}{2a}[/tex][tex]x_2 = \frac{-b - \sqrt{\Delta}}{2a}[/tex]In which:
[tex]\Delta = b^2 - 4ac[/tex]
The number of automobiles, in x years after 1980, is modeled by:
A(x) = 2x² + 24x + 300.
The amount is of 566,000 when A(x) = 566, hence the equation is:
2x² + 24x + 300 = 566
2x² + 24x - 266 = 0
x² + 12x - 133 = 0
The coefficients are a = 1, b = 12, c = -133, hence:
[tex]\Delta = 12^2 - 4(1)(-133) = 676[/tex][tex]x_1 = \frac{-12 + \sqrt{676}}{2} = 7[/tex][tex]x_2 = \frac{-12 - \sqrt{676}}{2} = -19[/tex]Time is a positive measure, hence we take the solution of 7, 1980 + 7 = 1987, which is the year.
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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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use yours formula to find the missing number of faces edges 15 vertices 9
Using Euler's Formula, the number of faces is given by: 8.
What does Euler's Formula states?It states that the number of vertices, edges and faces is related by the following equation:
V - E + F = 2.
In this problem, the parameters are given as follows:
E = 15, V = 9.
Hence the number of faces is given by:
V - E + F = 2.
9 - 15 + F = 2
F - 6 = 2
F = 8.
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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]
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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Please help me solve the problem in the image. I found that a=-65 and b=17 and when I plug it in the equation it’s wrong
Answer:
-181 - 79x
Step-by-step explanation:
I am not sure your comment fits the given problem.
I get the following approach and solution :
T(1 + 4x) = 1×a + 4x×b = 2 + 2x
a = 2 + 2x - 4xb
T(4 + 15x) = 4×a + 15x×b = -3 + 3x
4×(2 + 2x - 4xb) + 15xb = -3 + 3x
8 + 8x - 16xb + 15xb = -3 + 3x
11 + 5x - xb = 0
11 + 5x = xb
b = (11 + 5x)/x
a = 2 + 2x - 4x(11 + 5x)/x = 2 + 2x - 44 - 20x = -42 - 18x
T(3 - 5x) = 3×a - 5xb = 3(-42 - 18x) - 5x(11 + 5x)/x =
= -126 - 54x - 55 - 25x = -181 - 79x
control :
T(1 + 4x) = a + 4xb = -42 - 18x + 44 + 20x = 2 + 2x
T(4 + 15x) = 4a + 15xb = -168 - 72x + 165 + 75x = -3 + 3x
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)
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?
Learn more about the Slope here
https://brainly.com/question/3493733
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