All categories

3 years ago

HELP!!! you dont have to answer all the questions at least the first one!!!

Mathematics

1 answer:

UkoKoshka [18]3 years ago

7 0

57/19 = 3
x^5 / x^2 = x^(5-2) = x^3

so 57x^5 / 19x^2 = 3x^3

answer
3x^3

You might be interested in

Ms. Leeper gives her sixth- and seventh-grade students a geography test to decide who will represent her class in the school geo

goldfiish [28.3K]

C, the median is 23 when you add 13 + 10. Then you divide 23 and 2.

8 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

Starting on Day 1 with 1 jumping jack, Kiera doubles the number of jumping jacks she does every day. How many jumping jacks will

kolbaska11 [484]

Answer:

The answer is 512

Day 1:1

Day 2:2

Day 3:4

Day 4:8

Day 5:16

Day 6:32

Day 7:64

Day 8:128

Day 9:256

Day 10:512

5 0

3 years ago

Use the equation y = 7x to complete the table. What are the missing values of

storchak [24]

Answer:

Option D

Step-by-step explanation:

In this table, the constant of proportionality, which is just the number that each x value is multiplied by to get the corresponding y value, is 7. This is because, as shown by the first row of the table, 1*7=7, and, as shown by the third row of the table, 3*7=21. Therefore, in the second row of the table, y should be 14, as 2*7=14, while in the fourth row of the table, y should be 28, as 4*7=28.

Hope this helps! Comment any questions.

8 0

1 year ago

Read 2 more answers

What is the distance between (5,10) and (5,2)?

Licemer1 [7]

The distance is 8, since the x co-ordinate stays 5, the y co-ordinate goes from 10 to 2, which is an 8 distance

5 0

2 years ago