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2 years ago

Which expression is equivalent to 9s + 3 x 5 -2s?

2 answers:

6 0

The answer is A. When you take the 9s and the -2s, you get 7s. Then you add the 3x5 and end up with 15. Giving you the answer 7s+15

VARVARA [1.3K]2 years ago

3 0

Answer is A hope this helps

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Leokris [45]

Lamar used 80% of his data, so he used a greater percentage.

5 0

3 years ago

Read 2 more answers

What is the area of the kite?<br> 6 ft<br> 14 ft

Dovator [93]

Answer:

Step-by-step explanation:

if these are diagonals ,then

area=(product of diagonals)/2

=(6*14)/2=42 ft²

4 0

2 years ago

Tom,Scott and dawn sold cookies at the school fair.

aleksklad [387]

First of all work out how many tom and scott sold.
tom sold 60% of 90 cookies which is 54.
scott sold 2/3 of 150 cookies which is 100.
altogether the three of them sold 54+100+46 cookies which is 200.
Dawn sold 46 of these 200 cookies. To work out the percentage we have both the numbers to get 23 out of 100. We halved those numbers because a percentage is out of 100 we halved it to get it out of 100.
so 23 out of 100 is 23% so dawn sold 23%of the cookies.

Hope this helps :)

3 0

3 years ago

The price of a swimming pool has been discounted by 16.5%. The sale price is 1210.75. Find the original list price of the pool

yaroslaw [1]

To find the original pricing you must cross multiply
the answer would be 1450

5 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