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3 years ago

Can someone help me with this math problem?

Mathematics

1 answer:

skad [1K]3 years ago

3 0

$1\ mi-\dfrac{2}{3}\ mi=\dfrac{3}{3}\ mi-\dfrac{2}{3}\ mi=\dfrac{1}{3}\ m$

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Help. only questions 3 & 4 .

Maksim231197 [3]

I'm doing 3

For 3, using a table is very similar to a double number because the numbers are matching up in both ways.

On the bottom of a double number line we have like, for example,-- the bottom of the double number line would have batches. And its 1, 2, 3 ,4 , etc.

And on a table, it would be the same, the numbers on both diagrams have the same methods, have same way of lining things up but they're just drawn differently.

Hope this helped!

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3 years ago

Which one is bigger?<br>|-4| □ |-6|

Genrish500 [490]

Answer:

hello

Step-by-step explanation:

there is my answer

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3 years ago

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pentagon [3]

Answer:

Step-by-step explanation:

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3 years ago

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If y varies jointly with x and z, and y = 200 when x = 8 and z = 10, what is x when y = 165 and z = 11?

vova2212 [387]

Answer:

The answer is 2.5

The next blank is 6

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3 years ago

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If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v

pochemuha

Answer:

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = $5000(1 - \frac{1}{40}t )^2$ , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = $2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )$

V'(t) = $2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )$

When we simplify the above, we get

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0

4 years ago