Use the graph to write the explicit rule of the arithmetic sequence.
Question 19 options:

A)

ƒ(n) = 9 + 2(n – 1)

B)

ƒ(n) = 5 + 3(n – 1)

C)

ƒ(n) = –3 + 2(n – 1)

D)

ƒ(n) = 3 + 2(n – 1)

Combining of the two cylinders gives a composite figure that has the surface area of a solid cylinder of radius 7 mm and height 6 mm, such that the surface area of the figure is approximately 571.5 mm².

Radius of the inner cylinder, r = 3 mm

Radius of the outer cylinder, R = 7 mm

Height of the cylinders, h = 6 mm

Given that the inner cylinder exactly fits into the outer cylinder, we have;

The surface area of the composite figure is the surface area of the hollow outer cylinder + The area of the top and bottom of the inner cylinder.

Which gives;

The surface area of the composite figure, A, is the surface area of a solid cylinder, using the dimensions of the hollow outer cylinder.

Which gives;

Where, π = 3.14, we have;

A = 2 × 3.14 × 7² + 2 × 3.14 × 7 × 6 ≈ 571.5

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Answer:

a) 18

b) 48

c) 192

a) 30

b) 150

Step-by-step explanation:

Answer:

1

a ) 18

b ) 48

c ) 144

2

a ) 30

b ) 150..

[tex]...[/tex]

Answer:

C. 35.1

Step-by-step explanation:

First get the length of AC by using the Pythagorean Theorem, that ends up being about 11.18. Then you need to get the radius of the circle so you divide 11.18 by 2 and get 5.59. Then plug 5.59 for r into 2(pi)(r). 2(pi)(5.59) = 5.123

Answer:

Answer B is correct

Step-by-step explanation:

First, let us find the diameter (AC) of the circle using the Phythogerem theorem.

AC² = AB² + BC²

AC² = 2² + 11²

AC² = 4 + 121

AC² = 125

AC = √ 125

AC = 5√5

And now let us find the circumference of the circle.

C = π × d

C = π × 5√5

C = 5√5π in

The figure in the question is a box with dimensions of 6 in. by 2 in. by 9 in., which gives;

The dimensions of the box in the given figure are;

Length, L = 9 in.

Width, W = 2 in.

Height, H = 6 in.

The surface area, A, of a box is found using the following formula;

A = 2•L•W + 2•L•H + 2•H•W

Therefore;

A = 2×9×2 + 2×9×6 + 2×2×6 = 168

The volume of a box, V, is calculated as follows;

For the given figure, therefore;

V = 9 × 2 × 6 = 108

Cost of packaging per square inch, c = $0.004

Cost to package the figure, C, is therefore;

Which gives;

C = $0.004/in.² × 168 in.² = $0.672

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Answer: It is an equilateral square

Step-by-step explanation: it has 4 tick marks

Answer:

Step-by-step explanation:

15 hope it help

With the help of the area formulae of rectangles and triangles and the concept of surface area, the surface area of the composite figure is equal to 276 square centimeters.

The surface area of the truncated prism is the sum of the areas of its six faces, which are combinations of the areas of rectangles and right triangles. Then, we proceed to determine the surface area:

A = (12 cm) · (4 cm) + 2 · (3 cm) · (4 cm) + 2 · (12 cm) · (3 cm) + 2 · 0.5 · (12 cm) · (5 cm) + (5 cm) · (4 cm) + (13 cm) · (4 cm)

A = 48 cm² + 24 cm² + 72 cm² + 60 cm² + 20 cm² + 52 cm²

A = 276 cm²

With the help of the area formulae of rectangles and triangles and the concept of surface area, the surface area of the composite figure is equal to 276 square centimeters.

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Answer:

29, 21, 84

Step-by-step explanation:

Let the first number be x.

The second number is x - 8.

The third number is 4(x - 8).

Their sum is 134.

x + (x - 8) + 4(x - 8) = 134

x + 5(x - 8) = 134

x + 5x - 40 = 134

6x = 174

x = 29

x - 8 = 29 - 8 = 21

4(x - 8) = 4(29 - 8) = 4(21) = 84

The depth of the 84 inches wide satellite dish that has the receiver located 15 inches above the vertex is 29.4 inches.

The location of the receiver = 15 inches above the vertex

Width of the dish = 84 inches

The receiver of a satellite dish is normally located at the focus (focal point).

Taking the shape of the satellite dish as a parabola, and writing the equation of the parabola as x² = 4•a•y, we have;

Coordinates of the focus = (0, a)

Where;

a = The distance of the focus above the vertex

Therefore;

a = 15 inches

The dish is horizontally spaced equally about the vertex.

The farthest point horizontally from the vertex, X, of the dish corresponds to the furthest vertically from the vertex, which is the maximum depth or height, Y.

At the maximum height, from the vertex, Y, (farthest point, vertically from the vertex), we have;

X = 84 ÷ 2 = 42

Which gives;

42² = 4 × 15 × Y

Y = 42² ÷ (4 × 15) = 29.4

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The system of equations is -x + y = 4 and -2x + y = 0

To do this, we make use of the following ordered pairs

(4, 8)

The above point represents April 2008.

Let the system of equations be

mx + ny = 4

2mx + ny = 0

Substitute (4, 8) for x and y in the above equations.

4m + 8n = 4

8m + 8n = 0

Subtract both equations

4m - 8m = 4

This gives

-4m = 4

Divide by 4

m = -1

Substitute m = -1 in 8m + 8n = 0

-8 + 8n = 0

This gives

8n = 8

Divide by 8

n = 1

So, the system of equations is -x + y = 4 and -2x + y = 0

The system of equations is -x + y = 4 and -2x + y = 0

See attachment for the graph of the system of equation

The above is represented in the (a) part of this solution

Verify that the solution is (4, 8)

We have:

-x + y = 4 and -2x + y = 0

Substitute (4, 8) for x and y in the above equations.

-4 + 8 = 4 --- true

-2*4 + 8 = 0 --- true

Both equations are true

Hence, the system of equations have been verified

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The rate of interest applied on the amount is 15%.

According to the statement

we have given that the sum of money amounts to Rs. 2175 in 3 years and to Rs. 2625 in 5 years.

And we have to find the rate of interest on this money.

So, For this purpose,

The given equation is:

Money in 3 years = 2175

Money in 5 years = 2625

Amount increased in 2 years = 2625-2175

Amount increased in 2 years = 450

And amount increased per year = 450/2

Amount increased per year = 225.

SI in 3 years=450/2 *3=Rs. 675

Principal amount= 2175-675=1500

so again,

R=(I*100)/(P*T)

R=15%

So, The rate of interest applied on the amount is 15%.

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Answer:

x = 0, x = 5

Step-by-step explanation:

x² - 5x = 0 ← factor out x from each term

x(x - 5) = 0

equate each factor to zero and solve for x

x = 0

x - 5 = 0 ⇒ x = 5

If a quadrilateral exists as a parallelogram, then opposite sides exist congruent and parallel.

If a quadrilateral exists as a parallelogram, then opposite sides exist congruent and parallel.

Given: AB = 6cm , BC = 5cm, AD = 4cm, CD = 7cm and BD = 6cm​

Draw a line AB with 6cm with A as the center draw an arc of radius 4 cm. And draw another arc of 6cm to cut the previous arc at D with B as the center draw an arc of radius 5cm. And with D draw another arc of 7cm to cut the previous arc at C. Join the lines of the quadrilateral.

Now ABCD exists as a required quadrilateral.

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The area of quadrilateral ABCD exists equivalent to the area of triangle ABD plus the area of triangle ADC

Heron's Formula exists a technique for estimating the area of a triangle when you know the lengths of all three sides.

Therefore, the total area of  quadrilateral abcd exists 28.53 [tex]$$units^2[/tex].

Let a, b, and c be the lengths of the sides of a triangle.

The area is given by:

[tex]$$A = \sqrt{p(p-a)(p-b)(p-c)}[/tex]

where p exists half the perimeter

[tex]$p = \frac{a+b+c}{2}[/tex]

To estimate the area of triangle ABD

we have

a = AB = 2.89 units

b = BD = 8.59 units

c = DA = 8.6 units

To estimate the perimeter

[tex]$p = \frac{2.89+8.59+8.6}{2}[/tex]

= 10.4 units

To estimate the area

[tex]$A = \sqrt{10.04(10.04-2.89)(10.04-8.59)(10.04-8.6)}[/tex]

[tex]$A = \sqrt{10.04(7.15)(1.45)(1.44)}[/tex]

[tex]$A = \sqrt{149.89}[/tex]

A = 12.24 [tex]$$units^2[/tex]

To estimate the area of the triangle ADC

we have

a = AC = 4.3 units

b = AD = 8.6 units

c = DC = 7.58 units

To estimate the half perimeter p

[tex]$p = \frac{4.3+8.6+7.58}{2}[/tex]

= 10.24 units

To estimate the area

[tex]$A = \sqrt{10.24(10.24-4.3)(10.24-8.6)(10.24-7.58)}[/tex]

[tex]$A = \sqrt{10.24(5.94)(1.64)(2.66)}[/tex]

[tex]$A = \sqrt{265.35}[/tex]

A = 16.29 [tex]$$units^2[/tex]

To estimate the total area

A = 12.24+16.29 = 28.53 [tex]$$units^2[/tex].

Therefore, the total area of  quadrilateral abcd exists 28.53 [tex]$$units^2[/tex].

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Answer:

Below in bold.

Step-by-step explanation:

f(x) = x(18x + 4)

f(x) = 18x^2 + 4x

Antiderivative = 18 * (x^2+1)/(2 + 1) -+4x(1+1) / (1+1) + C

                        =  18x^3 / 3 + 4x^2 / 2 + C

                         = 6x^3 + 2x^2 + C.

Checking by differentiating:

f'(x) = 6*3 x^(3-1) + 2*2x^(2-1)

      =  18x^2 + 4x

      =  x(18 -+4)

The equation of the plane through the points p(0, 1, 1), q(1, 0, 1), and

r(1, 1, 0) is x + y + z = 2  .

According to the question

Let the points

p(0, 1, 1) = (x1,y1,z1)

q(1, 0, 1)= (x2,y2,z2)

r(1, 1, 0)= (x3,y3,z3)

now,

The equation of the plane through the points

i.e

The formula of equation of a plane passing through three non collinear points

[tex]\left[\begin{array}{ccc}x-x1&y-y1&z-z1\\x2-x1&y2-y1&z2-z1\\x3-x1&y3-y1&z3-z1\end{array}\right] = 0[/tex]

by substituting the values in the formula of equation of a plane

[tex]\left[\begin{array}{ccc}x-0&y-1&z-1\\1-0&0-1&1-1\\1-0&1-1&0-1\end{array}\right] = 0[/tex]

[tex]\left[\begin{array}{ccc}x&y-1&z-1\\1&-1&0\\1&0&-1\end{array}\right] = 0[/tex]  

x(1) - (y-1)(-1) + (z-1)(1) = 0

x + y - 1 + z - 1 = 0

x + y + z = 2

Hence, the equation of the plane through the points p(0, 1, 1), q(1, 0, 1), and r(1, 1, 0) is x + y + z = 2  .

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Considering the equation of the parabola, the coefficients are given as follows: p = 5, q = -20, r = 19.

The graph is the one given at the side of the question.

The equation of a quadratic function, of vertex (h,k), is given by:

y = a(x - h)² + k

In which a is the leading coefficient.

In this problem, the turning point is (2,-1), hence h = 2, k = -1, and the equation has the following format:

y = a(x - 2)² - 1

The curve passes through (3,4), that is, when x = 3, y = 4, and we use it to find a.

y = a(x - 2)² - 1

4 = a(3 - 2)² - 1

a = 5.

Hence the equation is:

y = 5(x - 2)² - 1

Placing it into standard form, we have that:

y = 5(x² - 4x + 4) - 1

y = 5x² - 20x + 19

Hence the coefficients are:

p = 5, q = -20, r = 19.

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Answer: 2000 meters

Step-by-step explanation: A kilometer means 1,000 meters a kilo means 1000. So, 2 x1000 = 2000 meters.

Answer:

X = -10

Step-by-step explanation:

2 = -8 -X

2+X = -8

X = -10

Answer:

x = - 10

Step-by-step explanation:

Comment

The trick to any equation like this one is to get the numbers on one side and the term with an unknown on the other side. That way, your last step is to divide by the number in front of the letter into a number on the other side.

Solution

2 = - 8 - x                                 Add 8 to both sides

2 + 8 = -8 + 8 - x                      Combine

10 = - x                                     The number in front of x is - 1. Divide.

10/-1 = -x/-1                        

-10 =x

The letter usually has to be expressed as a number greater than 0. In other words x is positive for this question

The restricted values in (x^2 - 3x - 2)/-4y * (2x - 2)/(-xy^2 + 2y^2) are x ≠2 and y ≠ 0

The complete question is added as an attachment

The equation is given as:

(x^2 - 3x - 2)/-4y * (2x - 2)/(-xy^2 + 2y^2)

Set the products of the denominators to 0

-4y * (-xy^2 + 2y^2) = 0

Split the equation

-4y = 0 or (-xy^2 + 2y^2) = 0

Remove the bracket

-4y = 0 or -xy^2 + 2y^2 = 0

Divide both sides by -4 in -4y = 0

y = 0

Factor out y^2 in -xy^2 + 2y^2 = 0

y^2(-x + 2) = 0

This means that y^2 = 0 or -x + 2= 0

This gives y = 0 or -x + 2 = 0

Solve for x in -x + 2 = 0

x = 2

Hence, the restricted values in (x^2 - 3x - 2)/-4y * (2x - 2)/(-xy^2 + 2y^2) are x ≠2 and y ≠ 0

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Into how many equal parts can the same cake be cut if the cuts can only be made along the gridlines, cake can be cut into 20 equal parts.

This is further explained below.

Generally, In spreadsheet software, gridlines are the light gray lines that divide the cells, rows, and columns. Gridlines are often used to maintain track of data. Gridlines are widely used in popular spreadsheet programs like Spreadsheet.

In conclusion, If the cuts can only be done along the gridlines, the cake may be divided into 20 equal pieces when cutting along the lines.

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Answer:

[tex]\sf \dfrac{4}{9}x^8[/tex]

Step-by-step explanation:

       [tex]\sf (a*b)^m = a^m *b^m\\\\(a^m)^n =a^{m*n}[/tex]

  16 = 4 *4 = 4²

 81  = 9 *9 = 9²

[tex]\sf\left (\dfrac{16}{81}x^{16}\right)^{\frac{1}{2}}= \left(\dfrac{4^2}{9^2}x^{16}\right)^{\frac{1}{2}}[/tex]

                [tex]\sf =\dfrac{4^{2*\frac{1}{2}}}{9^{2*\frac{1}{2}}}*x^{16*\frac{1}{2}}\\\\=\dfrac{4}{9}x^8[/tex]

Answer:

B

Step-by-step explanation:

Line A mostly follows the trend of the higher points, entirely neglecting the existence of the lower points. Line b acknowledges the lower points and adjusts the line accordingly to be lower.

Answer:

42

Step-by-step explanation:

One angle of a regular pentagon = 108°

One angle of a regular hexagon = 120°.

One angle of a square = 90°

∴ ∠BAC = 360 – (108 + 20 + 90) 360 – 318 = 42°

percentage of students say they have witnessed hazing and did not report it are  = 95%

No matter how voluntarily a person is to take part, hazing, initiation or deposition are all actions that are demanded of someone when they join or participate in a group and which humiliate, degrade, abuse, or imperil them.

involves humiliating a person or group. involves making fun of someone or something. involves or includes the deliberate destruction of or removal of private or public property for the purpose of affiliation with, initiation into, or as a requirement for continuing membership in an organization.

Paddling, beating, or other types of attack. Branding. consumption of disgusting substances or mixtures under duress or force.

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The solution of the questions is given as

a)

The alternative hypothesis is that mu is equal to 3.0. ( the average GPA at KSU is not equal to 3.0)

b)

This is a case with two separate arguments. (This is a two-tailed case)

c)

Test Statistic is given by:

z= 2.00

d)

Critical values of Z come from the table and equal [tex]\pm[/tex]1.96.

e)

, p-value = 0.0455

f)

CI= (3.006, 3.594)

g)

Conclusion: The evidence does not back up the assertion that KSU students have a cumulative grade point average of 3.0.

h)

Conclusion: The evidence does not back up the assertion that KSU students have a cumulative grade point average of 3.0.

i) Conclusion: The evidence does not lend credence to the assertion that KSU's typical grade point average is equivalent to 3.0.

j)

The choices you choose in Questions 7, 8, and 9 are identical.

a)

The null hypothesis states that mu equals 3.0. ( the average GPA at KSU is equal to 3.0) (Claim)

The alternative hypothesis is that mu is equal to 3.0. ( the average GPA at KSU is not equal to 3.0)

b)

This is a case with two separate arguments.

c)

Test Statistic is given by:

[tex]z=\frac{\barx-u^0}{σ/Vn}\\\\z=33 - 3.0/1.5/V100[/tex]

z= 2.00

d)

[tex]\alpha = 0.05[/tex]

Critical values of Z come from the table and equal [tex]\pm[/tex]1.96.

e)

Z Score = 2.00

reading from table, p-value = 0.0455

f)

Confidence Interval:

[tex]CI= \barx \pm \frac{ZX\sigma }{\sqrt((n)}[/tex]

[tex]CI =3.3 \pm 1.96 x 1.5 / \sqrt{100}[/tex]

CI= (3.006, 3.594)

g)

The disparity is noteworthy given that the computed value of Z = 2.00 is higher than the crucial value of z = 1.96. Cast doubt on the null hypothesis.

Conclusion: The evidence does not back up the assertion that KSU students have a cumulative grade point average of 3.0.

h)

The difference may be considered statistically significant since the p-value of 0.0455 is lower than the alpha threshold of 0.05. Cast doubt on the null hypothesis.

Conclusion: The evidence does not back up the assertion that KSU students have a cumulative grade point average of 3.0.

i)

The difference is statistically significant since the value of 3.0 is not included in the confidence range (3.006 - 3.594). Cast doubt on the null hypothesis.

Conclusion: The evidence does not lend credence to the assertion that KSU's typical grade point average is equivalent to 3.0.

j)

The choices you choose in Questions 7, 8, and 9 are identical.

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The answers are :

10) 25 : 24

11) 24 : 5

12) 36 : 5

13) 7 : 3

14) 15 : 1

Ratio = Number of event 1 : Number of event 2 (in same unit, if necessary)

10

Girls preferring orange juice : Boys preferring orange juice

50 : 48 (Divide by 2 on both sides)

25 : 24

11

Boys preferring orange juice : Boys preferring grapefruit juice

48 : 10 (Divide by 2 on both sides)

24 : 5

Remember :

12

3 minutes : 25 seconds

3 × 60 : 25 (Divide by 5 on both sides)

3 × 12 : 5

36 : 5

13

2 weeks : 6 days

2 × 7 : 6 (Divide by 2 on both sides)

7 : 3

14

5 hours : 20 minutes

5 × 60 : 20 (Divide by 20 on both sides)

5 × 3 : 1

15 : 1

In 90 minutes, she would have created 36 drawings.

The first step is to establish a pattern between time and the number of drawing that Frankita creates.

When the question is studied, it can be deduced that every 15 minutes the rate of increase in drawing is (previous drawing + (1 + rate of increase of previous drawing.

Drawing created in 30 minutes = 15 + 2 = 17

Drawing created in 45 minutes = 17 + ( 1 + 2) = 20

Drawing created in 60 minutes = 20 + (3 + 1 ) = 24

Drawing created in 75 minutes = 24 + (4 + 1) = 30

Drawing created in 90 minutes = 30 + (5 + 1) = 36

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In regression, the difference between the confidence interval and the prediction interval formula is "The addition of 1 to the quantity under the radical sign i.e., standard error".

The formula for the confidence interval is

[tex]y_h[/tex] ± [tex]t_{(1-\alpha /2,n-2)[/tex] × √(MSE([tex]\frac{1}{n}[/tex] + ([tex]x_h[/tex] - μ)²/∑([tex]x_i[/tex] - μ)²))

The formula for prediction interval is

Prediction interval = Sample estimate ± (t multiplier × standard error)

[tex]y_{new}[/tex] = [tex]y_h[/tex] ± [tex]t_{(1-\alpha /2,n-2)[/tex] × √(MSE(1 + [tex]\frac{1}{n}[/tex] + ([tex]x_h[/tex] - μ)²/∑([tex]x_i[/tex] - μ)²))

Where μ = x bar.

From the above, the prediction interval has one additional MSE term in the standard error calculation. But in the confidence interval, only one term is used.

So, the difference between them occurs in the standard error value.

The formula shows it by adding 1 to the quantity under the radical sign.

Thus, the difference is in the standard error.

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