X - Y =4 - 2x+y=18 Linnear system

Homework Jim Thompson! PART 2

BlackZzzverrR [31]

Problem 2

Part (a)

The 3D shape formed when rotating around the y axis forms a pencil tip

The shape formed when rotating around the x axis is a truncated cone turned on its side.

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Part (b)

Check out the two diagrams below.

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Problem 3

Answer: Choice A and Choice C

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Explanation:

Think of stacks of coins. Let's say we had 2 stacks of 10 quarters each. The quarters are identical, so they must produce identical volumes. Those sub-volumes then add up to the same volume for each stack. Now imagine one stack is perfectly aligned and the other stack is a bit crooked. Has the volume changed for the crooked stack? No, it hasn't. We're still dealing with the same amount of coins and they yield the same volume.

For more information, check out Cavalieri's Principle.

With all that in mind, this leads us to choice C. If the bases are the same, and so are the heights, then we must be dealing with the same volumes.

On the other hand, if one base is wider (while the heights are still equal) then the wider based block is going to have more volume. This leads us to choice A.

6 0

2 years ago

15 tan^3 x=5 tan x Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list. If

ANTONII [103]

$\bf 15tan^3(x)=5tan(x)\implies 3tan^3(x)=tan(x)
\\\\\\
3tan^3(x)-tan(x)=0\implies tan(x)[3tan^2(x)-1]=0
\\\\\\
\begin{cases}
tan(x)=0\\\\
\measuredangle x=0,\pi \\
----------\\
3tan^2(x)-1=0\\
tan^2(x)=\frac{1}{3}\\
tan(x)=\pm\sqrt{\frac{1}{3}}\\\\
tan(x)=\pm\frac{1}{\sqrt{3}}\\\\
\measuredangle x=\frac{\pi }{6}, \frac{5\pi }{6},\frac{7\pi }{6}, \frac{11\pi }{6}
\end{cases}$

8 0

3 years ago

No links please i cant seem to find the answer for this qn

Rudiy27

Answer:

(a) is 15

(b) is 30

75 divide by 15 times 90 divide by 15 is 30

6 0

3 years ago

Read 2 more answers

The area of the United States is 3,794,100 square miles. Which is the best estimation of this area?

olga55 [171]

Estimate
3,794,100
= 4,000,000
= 4 x 10^6

answer is C) 4 x 10^6

8 0

3 years ago

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

7 0

3 years ago