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3 years ago

6

Which expressions represent the distance between the two points?

2 answers:

vovikov84 [41]3 years ago

6 0

17 points between -15 and 2. Therefore, the correct equations are--

B. |–15 – 2|
&
<span>C. </span><span>|2 – (–15)|</span>

ICE Princess25 [194]3 years ago

6 0

<em>I am french, and I speak little english (I want to speak well), so please don't read my english errors...</em>

<u>Lesson :</u>

The distance between two points is an absolute value, like : $|a-b|$ (the distance between the point $a$ , and the point $b$ ).

For exemple, the distance between $(-2)$ and $19$ is $|-2+19| = |19-2| = 17$ .

<u>Your question :</u>

A. It's not a distance between two points because It's a negative result.
B. It's an absolute value, so it's positive.
C. It's an absolute value, so it's positive.
D. In mathematics, it doesn't mean anything.

So the distance between two points is B, and C.

Have a nice day !

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3 years ago

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Step-by-step explanation:

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Step-by-step explanation:

$\bigg( \huge{\frac{ {3}^{2}a ^{ - 2} }{3 {a}^{ - 1} }} \bigg)^{k} \\ \\ = \bigg( \huge{\frac{ 9a }{3 {a}^{ 2} }} \bigg)^{k} \\ \\ = \bigg( \huge{\frac{ 3 }{ {a}}} \bigg)^{k} \\ \\ = \bigg( \huge{\frac{ 3 }{ {5}}} \bigg)^{ - 2} \\ \\ = \bigg( \huge{\frac{ 5 }{ {3}}} \bigg)^{ 2} \\ \\ = \huge{\frac{ 25 }{ {9}}}$

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4 years ago

Read 2 more answers

A conical water tank with vertex down has a radius of 13 feet at the top and is 21 feet high. If water flows into the tank at a

VLD [36.1K]

Answer:

$\frac{dh}{dt}\approx0.08622\text{ ft/min}$

Step-by-step explanation:

We know that the conical water tank has a radius of 13 feet and is 21 feet high.

We also know that water is flowing into the tank at a rate of 30ft³/min. In other words, our derivative of the volume with respect to time t is:

$\frac{dV}{dt}=\frac{30\text{ ft}^3}{\text{min}}$

We want to find how fast the depth of the water is increasing when the water is 17 feet deep. So, we want to find dh/dt.

First, remember that the volume for a cone is given by the formula:

$V=\frac{1}{3}\pi r^2h$

We want to find dh/dt. So, let's take the derivative of both sides with respect to the time t. However, first, let's put the equation in terms of h.

We can see that we have two similar triangles. So, we can write the following proportion:

$\frac{r}{h}=\frac{13}{21}$

Multiply both sides by h:

$r=\frac{13}{21}h$

So, let's substitute this in r:

$V=\frac{1}{3}\pi (\frac{13}{21}h)^2h$

Square:

$V=\frac{1}{3}\pi (\frac{169}{441}h^2)h$

Simplify:

$V=\frac{169}{1323}\pi h^3$

Now, let's take the derivative of both sides with respect to t:

$\frac{d}{dt}[V]=\frac{d}{dt}[\frac{169}{1323}\pi h^3}]$

Simplify:

$\frac{dV}{dt}=\frac{169}{1323}\pi \frac{d}{dt}[h^3}]$

Differentiate implicitly. This yields:

$\frac{dV}{dt}=\frac{169}{1323}\pi (3h^2)\frac{dh}{dt}$

We want to find dh/dt when the water is 17 feet deep. So, let's substitute 17 for h. Also, let's substitute 30 for dV/dt. This yields:

$30=\frac{169}{1323}\pi (3(17)^2)\frac{dh}{dt}$

Evaluate:

$30=\frac{146523}{1323}\pi( \frac{dh}{dt})$

Multiply both sides by 1323:

$39690=146523\pi\frac{dh}{dt}$

Solve for dh/dt:

$\frac{dh}{dt}=\frac{39690}{146523}\pi$

Use a calculator. So:

$\frac{dh}{dt}\approx0.08622\text{ ft/min}$

The water is rising at a rate of approximately 0.086 feet per minute.

And we're done!

Edit: Forgot the picture :)

3 0

3 years ago

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rjkz [21]

Answer: B. 4.5 X 0.72

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3 years ago