Answer:
33%
Step-by-step explanation:
Why it is 33% is you minus 20 and 30 to get 10 then you just put that on top of 30 to get 10/30 and you will get 33%.
Answer:
The answer is (d) ⇒ ![pq^{2}r\sqrt[3]{pr^{2}}](https://tex.z-dn.net/?f=pq%5E%7B2%7Dr%5Csqrt%5B3%5D%7Bpr%5E%7B2%7D%7D)
Step-by-step explanation:
* To simplify the cube roots:
If its number then the number must be written in the form x³
then we divide the power by 3 to cancel the radical
If its variable we divide its power by 3 to cancel the radical
∵ ![\sqrt[3]{p^{4}q^{6}r^{5}}=p^{\frac{4}{3}}q^{\frac{6}{3}}r^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bp%5E%7B4%7Dq%5E%7B6%7Dr%5E%7B5%7D%7D%3Dp%5E%7B%5Cfrac%7B4%7D%7B3%7D%7Dq%5E%7B%5Cfrac%7B6%7D%7B3%7D%7Dr%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
∴ 
∵ ![p^{\frac{1}{3}}=\sqrt[3]{p}](https://tex.z-dn.net/?f=p%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7Bp%7D)
∵ ![r^{\frac{2}{3}}=\sqrt[3]{r^{2}}](https://tex.z-dn.net/?f=r%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7Br%5E%7B2%7D%7D)
∴ ![p(p)^{\frac{1}{3}}q^{2}r(r)^{\frac{2}{3}}=p(\sqrt[3]{p})q^{2}r(\sqrt[3]{r^{2}})](https://tex.z-dn.net/?f=p%28p%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7Dq%5E%7B2%7Dr%28r%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3Dp%28%5Csqrt%5B3%5D%7Bp%7D%29q%5E%7B2%7Dr%28%5Csqrt%5B3%5D%7Br%5E%7B2%7D%7D%29)
∴ ![prq^{2}\sqrt[3]{pr^{2}}}](https://tex.z-dn.net/?f=prq%5E%7B2%7D%5Csqrt%5B3%5D%7Bpr%5E%7B2%7D%7D%7D)
∴ The answer is (d)
Answer: - 1 3/4
Step-by-step explanation: You have to keep the first number and flip the second number, and you end with -1 3/4
PLEASE MARK BRAINLIEST
<span>The sign value for x is positive and for y is negative. In standard position, the x point is at the origin, so the value will be 0. The terminal side is the line that moves to create the angle. At -60 degrees, the terminal side will be in the minus side of the y axis.</span>
The slope of this line is
−
12
5
.
Perpendicular lines have opposite reciprocal slopes. To find an opposite reciprocal, multiply by
−
1
and flip the numerator and denominator.
Hence the opposite reciprocal of
−
12
5
and the slope of a line perpendicular to a line with a slope of
−
12
5
is
5
12
.
