Answer:
The distance of the point from the origin = 9.29 units.
Step-by-step explanation:
Given point:
(7,-6)
The angle lies such that the terminal side of the angle contains the given point.
To draw the angle and find the distance from the origin to the given point.
Solution:
The terminal side of the angle is where the angle ends with the initial side being the positive side of the x-axis.
So, we can plot the point (7,-6) by moving 7 units on the x-axis horizontally and -6 units on the y-axis vertically.
We can find the distance of the point from the origin by find the hypotenuse of the triangle formed.
Applying Pythagorean theorem.
![Hypotenuse^2=Shorter\ Leg^2+Shortest\ Leg^2](https://tex.z-dn.net/?f=Hypotenuse%5E2%3DShorter%5C%20Leg%5E2%2BShortest%5C%20Leg%5E2)
![Hypotenuse^2 = (7)^2+(-6)^2](https://tex.z-dn.net/?f=Hypotenuse%5E2%20%3D%20%287%29%5E2%2B%28-6%29%5E2)
![Hypotenuse^2=49+36\\Hypotenuse^2=85](https://tex.z-dn.net/?f=Hypotenuse%5E2%3D49%2B36%5C%5CHypotenuse%5E2%3D85)
Taking square root both sides :
![\sqrt{Hypotenuse^2}=\sqrt{85}](https://tex.z-dn.net/?f=%5Csqrt%7BHypotenuse%5E2%7D%3D%5Csqrt%7B85%7D)
![Hypotenuse = 9.29\ units](https://tex.z-dn.net/?f=Hypotenuse%20%3D%209.29%5C%20units)
Thus, the distance of the point from the origin = 9.29 units.
The figure is shown below.
Answer:
2408.55 cubic yards
Step-by-step explanation:
use pi times radius squared times height, then divide by 3
SO SORRY I ANSWERED ON THE OTHER ONE! The answer is C. This line has a negative slope, and the ending number tells you the y-intercept.
Answer: The distance around the lot is given by
![4\frac{1}{20}\ miles](https://tex.z-dn.net/?f=4%5Cfrac%7B1%7D%7B20%7D%5C%20miles)
Step-by-step explanation:
Since we have given that
Length of rectangular lot is given by
![1\frac{1}{8}\\\\=\frac{9}{8}\ miles](https://tex.z-dn.net/?f=1%5Cfrac%7B1%7D%7B8%7D%5C%5C%5C%5C%3D%5Cfrac%7B9%7D%7B8%7D%5C%20miles)
Width of rectangular lot is given by
![\frac{9}{10}\ mile](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B10%7D%5C%20mile)
We need to find the distance around the lot.
As we know the formula for "Perimeter of rectangle":
![Perimeter=2(Length+Width)\\\\Perimeter=2(\frac{9}{8}+\frac{9}{10})\\\\Perimeter=2(\frac{45+36}{40})\\\\Perimeter=2\times \frac{81}{40}\\\\Perimeter=\frac{81}{20}\\\\Perimeter=4\frac{1}{20}\ miles](https://tex.z-dn.net/?f=Perimeter%3D2%28Length%2BWidth%29%5C%5C%5C%5CPerimeter%3D2%28%5Cfrac%7B9%7D%7B8%7D%2B%5Cfrac%7B9%7D%7B10%7D%29%5C%5C%5C%5CPerimeter%3D2%28%5Cfrac%7B45%2B36%7D%7B40%7D%29%5C%5C%5C%5CPerimeter%3D2%5Ctimes%20%5Cfrac%7B81%7D%7B40%7D%5C%5C%5C%5CPerimeter%3D%5Cfrac%7B81%7D%7B20%7D%5C%5C%5C%5CPerimeter%3D4%5Cfrac%7B1%7D%7B20%7D%5C%20miles)
Hence, the distance around the lot is given by
![4\frac{1}{20}\ miles](https://tex.z-dn.net/?f=4%5Cfrac%7B1%7D%7B20%7D%5C%20miles)