Answer:
24 years
Step-by-step explanation:
total ratio =8
older student=40 years
3/8*40 ÷ 5/8=24
As per the the question we need to calculate the value of first 4 multiplied by 10.
Now in the number 4,043, the first 4 is at the thousands place. So the place value of first 4 in the digit 4,043 is 4000.
Multiplying the value of first 4 by 10 we get:

Now, the second 4 in the digit 4,043 is at the tens place. So the place value of second 4 in the digit 4,043 is 40.
We can see that two of them are not equal and the value of first 4 times ten is far greater than the value of second 4.
It can be said that two of the 4s in the digit 4,043 are at two different places, first is at thousands place and second is at tens place. So if the numbers are two places apart their place value will have difference of 100 folds.
Answer:

Step-by-step explanation:
In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).
The formula we will use for this problem is the following:

where:


a=0

so the volume becomes:

This can be simplified to:

and the integral can be rewritten like this:

which is a standard integral so we solve it to:
![V=9\pi[tan y]\limits^\frac{\pi}{3}_0](https://tex.z-dn.net/?f=V%3D9%5Cpi%5Btan%20y%5D%5Climits%5E%5Cfrac%7B%5Cpi%7D%7B3%7D_0)
so we get:
![V=9\pi[tan \frac{\pi}{3} - tan 0]](https://tex.z-dn.net/?f=V%3D9%5Cpi%5Btan%20%5Cfrac%7B%5Cpi%7D%7B3%7D%20-%20tan%200%5D)
which yields:
]
The graph will cross at the coordinates (-2, 9)
<h3>How to solve equations?</h3>
y = 3x + 15
y = 3 - 3x
y = 3x + 15
Hence,
when x = 2
y = 3(2) + 15 = 21
when x = 0
y = 3(0) + 15 = 15
y = 3 - 3x
when x = 2
y = 3 - 3(2)
y = 3 - 6
y = -3
when x = 0
y = 3 - 3(0)
y = 3
Therefore, let's check if the equation will cross.
y = 3x + 15
y = 3 - 3x
using substitution,
3 - 3x = 3x + 15
3 - 15 = 3x + 3x
- 12 = 6x
x = -12 / 6
x = -2
y = 3 - 3(-2)
y = 3 + 6
y = 9
Therefore, the graph will cross at the coordinates (-2, 9)
learn more on equations here: brainly.com/question/19297665
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