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3 years ago

Given that a^b = x, evaluate the following: 2a^3b +a^2b+ a^b

Mathematics

1 answer:

Bingel [31]3 years ago

8 0

Answer:

2x^3+x^2 +x = x(2x^2 +x+1)

Step-by-step explanation:

a^b = x,

2a^3b +a^2b+ a^b=

2(a^b)^3+(a^b)^2 +a^b=

2x^3+x^2 +x=

x(2x^2 +x+1)

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Pat needs 14 sticks of pepperoni to make 28 pizzas. How many sticks of pepperoni are needed to make 36 pizzas?

VladimirAG [237]

Answer:

Step-by-step explanation:

It is 18 because 14 x 2 equal 28 so that would mean you need one stick to make 2 pizzas. Then you divide 36 by 2 and get 18

4 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

A veterinarian estimated the weight of a puppy to be 7 kg. The actual weight of the puppy was 6.6 kg.

KIM [24]

9514 1404 393

Answer:

absolute error: 0.4 kg
relative error: 6.1%

Step-by-step explanation:

The absolute error is the difference between the estimate and the actual weight:

7.0 kg -6.6 kg = 0.4 kg . . . . absolute error

The relative error is the ratio of the absolute error to the actual weight:

(0.4 kg)/(6.6 kg) × 100% = 6.060606...% ≈ 6.1% . . . . relative error

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3 years ago

Shiva brought $39.00 to the state fair. She bought a burger, a souvenir, and a pass. The burger was 1/3 as much as the souvenir,

777dan777 [17]

Pass = $21 , Souvenir = 2x/3 = $14 , Burger = x/6 =$3.5 .

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3 years ago

a seven digit number has 0 in the ones place , a 6 in the ten thousands place and an 8 in the millions place and fives in each o

PSYCHO15rus [73]

Here ya go friend
8565550

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4 years ago

Read 2 more answers

Given that a^­­b = x, evaluate the following: 2a^3b +a^2b+ a^b

Given that a^b = x, evaluate the following: 2a^3b +a^2b+ a^b