The slope of a line is always zero because the line does not move up or down on the y-axis.
Answer:
the distance
Step-by-step explanation:
More time care travels, more distance it goes. So distance depends of the time.
Do you mean
![3 \sqrt[3]{125}](https://tex.z-dn.net/?f=3%20%5Csqrt%5B3%5D%7B125%7D%20)
or
![\sqrt[3]{125}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%20)
remember that
![\sqrt[n]{x^m}=x^\frac{m}{n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Em%7D%3Dx%5E%5Cfrac%7Bm%7D%7Bn%7D)
resolve the
![\sqrt[3]{125}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D)
part first
![\sqrt[3]{125}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%3D)
![\sqrt[3]{5^3}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5%5E3%7D%3D)
so
![3 \sqrt[3]{125}=3*5=15](https://tex.z-dn.net/?f=3%20%5Csqrt%5B3%5D%7B125%7D%3D3%2A5%3D15%20)
or
![\sqrt[3]{125} =5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%20%3D5)
not sure which one you mean
9514 1404 393
Answer:
2√30 ∠-120°
Step-by-step explanation:
The modulus is ...
√((-√30)² +(-3√10)²) = √(30 +90) = √120 = 2√30
The argument is ...
arctan(-3√10/-√30) = arctan(√3) = -120° . . . . a 3rd-quadrant angle
The polar form of the number can be written as ...
(2√30)∠-120°
_____
<em>Additional comments</em>
Any of a number of other formats can be used, including ...
(2√30)cis(-120°)
(2√30; -120°)
(2√30; -2π/3)
2√30·e^(i4π/3)
Of course, the angle -120° (-2π/3 radians) is the same as 240° (4π/3 radians).
__
At least one app I use differentiates between (x, y) and (r; θ) by the use of a semicolon to separate the modulus and argument of polar form coordinates. I find that useful, as a pair of numbers (10.95, 4.19) by itself does not convey the fact that it represents polar coordinates. As you may have guessed, my personal preference is for the notation 10.95∠4.19. (The lack of a ° symbol indicates the angle is in radians.)
3y+21+8y
11y+21 is the answer
