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3 years ago

Tyler is overdrawn at the bank by $180.His brother has $70 more than him.How much money does lyler brother have?

Mathematics

1 answer:

____ [38]3 years ago

8 0

Answer: - $110

Step-by-step explanation: If Tyler is overdrawn at the bank it means that he has a negative balance. Tylers balance is -$180. If his brother has $70 more than him you need to add the two numbers together.

-$180 + $70 = -$110

Even though Tyler’s brother has more money than Tyler, his brother is still overdrawn by $110. (-$110)

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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

I need helpp !!!!!!!!!!!

natta225 [31]

Answer:

In descending order, 1, 4, 3

4 0

3 years ago

Write each number as a logarithm with base 2:-3

expeople1 [14]

The number -3 written as a logarithm with a base of 2 is log₂(0.125) or log₂(1/8)

<h3>What are logarithms?</h3>

As a general rule, logarithms are mathematical expressions that are written in the form log(x) or ln(x), for natural logarithms

<h3>How to rewrite the number as a logarithm?</h3>

The number is given as:

x = -3

The base of the logarithm is given as:

Base = 2

To rewrite the given number as a base of 2, we take the exponent of the number where the base is 2

This is represented as:

Number =2^-3

Apply the power rule of indices

Number =1/2^3

Evaluate the exponent

Number = 1/8

Evaluate the quotient

Number = 0.125

Hence, when the number -3 is rewritten as a logarithm with base 2, the equivalent logarithm expression is log₂(0.125) or log₂(1/8)