Since 6/3=2, we could make it 2-y-2/y. If we wanted everything over y, we could make it (2y-y^2-2)/y by multiplying both 2 and y with y. This is not factorable due to that nothing multiplies to -2 and adds up to 2.
Answer:
Number line D is true
Step-by-step explanation:
Given
![-3x + 1 < 7](https://tex.z-dn.net/?f=-3x%20%2B%201%20%3C%207)
See attachment for number lines
Required
Which number line makes the inequality true
First, we solve for x in ![-3x + 1 < 7](https://tex.z-dn.net/?f=-3x%20%2B%201%20%3C%207)
![-3x < 7 - 1](https://tex.z-dn.net/?f=-3x%20%3C%207%20-%201)
![-3x < 6](https://tex.z-dn.net/?f=-3x%20%3C%206)
Divide through by -3
![\frac{-3x}{-3} < \frac{6}{-3}](https://tex.z-dn.net/?f=%5Cfrac%7B-3x%7D%7B-3%7D%20%3C%20%5Cfrac%7B6%7D%7B-3%7D)
![x > -2](https://tex.z-dn.net/?f=x%20%3E%20-2)
The inequality means x is greater than -2.
This means that, the line must point to the right of -2 and the "greater than" means that the circle must be opened
Number line D is true
Answer:
The radius of the spool is 3.7 cm
Step-by-step explanation:
Consider the provided information.
The rate is 59.4 cm per sec and it makes 152 revolutions per min.
We need to find the radius.
Now use the formula: ![r=\frac{v}{\omega}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bv%7D%7B%5Comega%7D)
Substitute
in above formula and change the revolutions per min to rad per second.
![r=59.4\frac{cm}{sec}(\frac{1}{152}\times \frac{min}{rev}\times\frac{60sec}{1min}\times\frac{1rev}{2\pi rad})](https://tex.z-dn.net/?f=r%3D59.4%5Cfrac%7Bcm%7D%7Bsec%7D%28%5Cfrac%7B1%7D%7B152%7D%5Ctimes%20%5Cfrac%7Bmin%7D%7Brev%7D%5Ctimes%5Cfrac%7B60sec%7D%7B1min%7D%5Ctimes%5Cfrac%7B1rev%7D%7B2%5Cpi%20rad%7D%29)
![r=\frac{3564}{304\pi}cm](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B3564%7D%7B304%5Cpi%7Dcm)
![r\approx 3.7cm](https://tex.z-dn.net/?f=r%5Capprox%203.7cm)
Hence, the radius of the spool is 3.7 cm