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3 years ago

Will mark as brainliest if correct

Mathematics

1 answer:

Dvinal [7]3 years ago

5 0

5= 7 ^x
X=ln5/ln7
X=0.827

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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

Michelle opened a savings account for her emergency fund. She deposited $2,000 into her account, which earns 2.10% interest, com

mrs_skeptik [129]

Answer: she will have $2042.4 have in the account after 1 year.

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $2000

r = 2.1% = 2.1/100 = 0.021

n = 12 because it was compounded 12 times in a year.

t = 1 year

Therefore,

A = 2000(1 + 0.021/12)^12 × 1

A = 2000(1 + 0.00175)^12

A = 2000(1.00175)^12

A = $2042.4

5 0

3 years ago

What is the following product? (2V7 +3V6)(5V2+4V3)

raketka [301]

$(2\sqrt{7}+3\sqrt{6})(5\sqrt{2}+4\sqrt{3}) \\\\\\ \begin{array}{lllllll} 10\sqrt{14}&+&15\sqrt{12}&+&8\sqrt{21}&+&12\sqrt{18}\\\\ &&15\sqrt{2^2\cdot 3}&&&&12\sqrt{3^2\cdot 2}\\\\ &&15\cdot 2\sqrt{3}&&&&12\cdot 3\sqrt{2}\\\\ &&30\sqrt{3}&&&&36\sqrt{2}\\\\ 10\sqrt{14}&+&30\sqrt{3}&+&8\sqrt{21}&+&36\sqrt{2}\\\\ 10\sqrt{14}&+&8\sqrt{21}&+&30\sqrt{3}&+&36\sqrt{2} \end{array}$