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2 years ago

11

WORTH 6 POINTSSomeone please help me. this is due tommorow.

1 answer:

tiny-mole [99]2 years ago

8 0

Here are a few rules that apply to this expression:

Powering a power: $(x^m)^n=x^{m*n}$
Converting a negative exponent to a positive one: $x^{-m}=\frac{1}{x^m};\frac{1}{x^{-m}}=x^m$

Firstly, solve the exponent: $(-6x)^{-2}=-6^{-2}x^{1*-2}=-6^{-2}x^{-2}$

Next, convert the negative exponents: $-6^{-2}x^{-2}=-\frac{1}{6^2x^2}=-\frac{1}{36x^2}$

<u>Your final answer is -1/36x^2.</u>

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If you have a lowest score of 21 and a range of 47, your highest score will be:________

aalyn [17]

Answer: 68

Step-by-step explanation:

range=47 lowest score = 21 let highest score =x

Range=highest score - lowest score

47 = x - 21

x = 47 + 21

x = 68

Therefore the highest score is 68

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2 years ago

Nico did 42 push-ups in 3 minutes.

Luda [366]

Answer:

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Step-by-step explanation:

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2 years ago

Help pleasee<br> On khan academy under geometry unit test.

Rzqust [24]

X=147
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5 0

2 years ago

Read 2 more answers

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

Can someone please help me with number 2 please I’ll love you forever ;)

yawa3891 [41]

Step-by-step explanation:

the first question

x²+x+5

when x is -4

(-4)²+(-4)+5

16-4+5

12+5

17

hope this was helpful

7 0

2 years ago