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3 years ago

Billy watched the clown car drive up to the circus and saw

Mathematics

1 answer:

nekit [7.7K]3 years ago

6 0

14%=56%
56%=56/100=0.56

14=0.56 of all
14=0.56 times all
divide both sides by 0.56
25=all

25 performers total

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What is the area of the figure?

son4ous [18]

Yes. 95 is correct.
You have three congruent "indentations" in the right hand side. Thus each section must be 15/3 = 5 cm long.
The top rectangle will be 8*5 = 40 cm^2
The bottom rectangle will also be 8*5 = 40 cm^2
The middle area will be 5(8-5) = 5 * 3 = 15 cm^2
40 + 40 + 15 = 95

5 0

3 years ago

82. Subtract 12x? + 6y - 3 and 3x? - 9y + 12

balu736 [363]

Answer:

-x^2+15y-15

Answer if it is a ^2

Step-by-step explanation:

2x^2 + 6y - 3 -(3x^2 - 9y + 12)=

2x^2+6y-3-3x^2+9y-12=

-x^2+15y-15

is it a ^2 or a question mark

3 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

Sophia sells makeup brushes for b dollars. For

mel-nik [20]

C.0.25
Meaning 25% 25 divide one hundred
Therefore answer is 0.25

5 0

2 years ago

20 POINTS!!!!!

inn [45]

Answer:

D'accord. Ainsi, la flèche ou le pointeur doit pointer dessus. Si ne pointe pas, que. Découvrez-le

7 0

2 years ago