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3 years ago

Drag and drop the correct volume of the cylinder into the box to find the volume of the

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

8 0

Answer:

1491.5

Step-by-step explanation:

radius is 5in so pi * 5 squared * 19 will get you this answer

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Please help me with this question

netineya [11]

Answer

777

Step-by-step explanation:

jnimimimimimim000

3 0

3 years ago

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On the day of a child’s birth, a deposit of $25,000 is made in a trust fund that pays 8.75% interest, compounded continuously. D

lbvjy [14]

9514 1404 393

Answer:

$222,822.57

Step-by-step explanation:

The account balance is given by ...

A = P·e^(rt)

where P is the principal invested at interest rate r for t years.

A = $25,000×e^(0.0875·25) ≈ $222,822.57

8 0

3 years ago

What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S

Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is $y=\frac{1}{3} x+2$

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3

And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.

Let required slope = x

$\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}$

So we need to find the equation of a line whose slope is $\frac{1}{3}$ and passing through (-3,1)

Equation of line passing through $(x_1 , y_1)$ and having lope of m is given by

$\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)$

$\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}$

Substituting the values we get,

$\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}$