Range, Maximum, and y-intercept
Answer:
![\sqrt[3]{a^{2}+b^{2}}=(a^{2}+b^{2})^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Ba%5E%7B2%7D%2Bb%5E%7B2%7D%7D%3D%28a%5E%7B2%7D%2Bb%5E%7B2%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
Step-by-step explanation:
∵∛x = (x)^1/3
∴
So you can replace the radicals by fractional exponents
Answer:
5/7
Step-by-step explanation:
Looks fishy to me, the way this problem combines <, = and > symbols.
But anyway. Subtract 3x from both sides. You'll get
5 < x = 2 > 0 2 > 0 is true, but that's a different type of math problem (true or false). If you decide to substitute '1' for 2 > 0, then the "1" says "TRUE."
Then 5 < x = true, so you must find x values that are larger than 5.
That would be (5, infinity)
Please go back and ensure that you have copied down the original problem exactly as it appears.
Answer:
y = 1/3x + 7
Step-by-step explanation:
Perpendicular slope is opposite reciprocal so it will be 1/3
Use the slope and point given and plug into y = mx + b
4 = 1/3 (-9) + b
4 = -3 + b
7 = b
write new equation with new b and m
y = 1/3x + 7
