Answer:
answer::::: 9 . Is it true ?
the new radius be to meet the client's need is 4.9 cm .
<u>Step-by-step explanation:</u>
Here we have , can company makes a cylindrical can that has a radius of 6 cm and a height of 10 cm. One of the company's clients needs a cylindrical can that has the same volume but is 15 cm tall. We need to find What must the new radius be to meet the client's need . Let's find out:
Let we have two cylinders of volume
with parameters as follows :
![r_1=6cm\\h_1=10cm\\r_2=?\\h_2=15cm](https://tex.z-dn.net/?f=r_1%3D6cm%5C%5Ch_1%3D10cm%5C%5Cr_2%3D%3F%5C%5Ch_2%3D15cm)
We know that volume of cylinder is
, According to question volume of both cylinder is equal i.e
⇒ ![V_1=V_2](https://tex.z-dn.net/?f=V_1%3DV_2)
⇒ ![\pi (r_1)^2h_1= \pi (r_2)^2h_2](https://tex.z-dn.net/?f=%5Cpi%20%28r_1%29%5E2h_1%3D%20%5Cpi%20%28r_2%29%5E2h_2)
⇒ ![(r_1)^2h_1= (r_2)^2h_2](https://tex.z-dn.net/?f=%28r_1%29%5E2h_1%3D%20%28r_2%29%5E2h_2)
⇒ ![\frac{(r_1)^2h_1}{h_2}= (r_2)^2](https://tex.z-dn.net/?f=%5Cfrac%7B%28r_1%29%5E2h_1%7D%7Bh_2%7D%3D%20%28r_2%29%5E2)
⇒
Putting all values
⇒ ![(r_2) =\sqrt{ \frac{(6)^2(10)}{15}}](https://tex.z-dn.net/?f=%28r_2%29%20%3D%5Csqrt%7B%20%5Cfrac%7B%286%29%5E2%2810%29%7D%7B15%7D%7D)
⇒ ![(r_2) =\sqrt{ \frac{36(10)}{15}}](https://tex.z-dn.net/?f=%28r_2%29%20%3D%5Csqrt%7B%20%5Cfrac%7B36%2810%29%7D%7B15%7D%7D)
⇒ ![(r_2) =\sqrt{ \frac{360}{15}}](https://tex.z-dn.net/?f=%28r_2%29%20%3D%5Csqrt%7B%20%5Cfrac%7B360%7D%7B15%7D%7D)
⇒ ![(r_2) =\sqrt{24}](https://tex.z-dn.net/?f=%28r_2%29%20%3D%5Csqrt%7B24%7D)
⇒ ![(r_2) =4.9cm](https://tex.z-dn.net/?f=%28r_2%29%20%3D4.9cm)
Therefore , the new radius be to meet the client's need is 4.9 cm .
Answer:
1. x = 2
2. x = 6
3. x = 6
4. x = 1
Step-by-step explanation: i think thats it
Answer:
First answer: x=1/2y+−3/2
Second answer: x=−1/4y+9/4
Step-by-step explanation: hope this help
Answer:
which solution are you looking for?
Step-by-step explanation:
Solving for X would be: x<0
Graph inequality would be the left side shaded i think