Add 1/4 to both sides to get T=-2/4
Answer: Suppose you're given the two points (–2, 1) and (1, 5), and they want you to find out how far apart they are. You can draw in the lines that form a right-angled triangle, using these points as two of the corners. It's easy to find the lengths of the horizontal and vertical sides of the right triangle: just subtract the x-values and the y-values.
Step-by-step explanation: Then use the Pythagorean Theorem to find the length of the third side (which is the hypotenuse of the right triangle):
c2 = a2 + b2
...so:
c
2
=(5−1)
2
+(1−(−2))
2
c = \sqrt{(5 - 1)^2 + (1 - (-2))^2}c=
(5−1)
2
+(1−(−2))
2
c = \sqrt{(4)^2 + (3)^2}c=
(4)
2
+(3)
2
c = \sqrt{16 + 9} = \sqrt{25} = 5c=
16+9
=
25
=5
Answer:
![y=-2.18,-0.22](https://tex.z-dn.net/?f=y%3D-2.18%2C-0.22)
Step-by-step explanation:
We have been given the equation
![(5y+6)^2=24](https://tex.z-dn.net/?f=%285y%2B6%29%5E2%3D24)
Take square root both sides. When ever we take square root, we include plus minus sign
![\sqrt{(5y+6)^2}=\pm\sqrt{24}\\\\5y+6=\pm2\sqrt6](https://tex.z-dn.net/?f=%5Csqrt%7B%285y%2B6%29%5E2%7D%3D%5Cpm%5Csqrt%7B24%7D%5C%5C%5C%5C5y%2B6%3D%5Cpm2%5Csqrt6)
Now, subtract 6 to both sides of the equation
![5y=-6\pm2\sqrt6](https://tex.z-dn.net/?f=5y%3D-6%5Cpm2%5Csqrt6)
Divide both sides by 5
![y=\frac{1}{5}(-6\pm2\sqrt6)](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B5%7D%28-6%5Cpm2%5Csqrt6%29)
![y=\frac{1}{5}(-6+2\sqrt6),y=\frac{1}{5}(-6-2\sqrt6)](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B5%7D%28-6%2B2%5Csqrt6%29%2Cy%3D%5Cfrac%7B1%7D%7B5%7D%28-6-2%5Csqrt6%29)
In decimal we have
![y=-2.18,-0.22](https://tex.z-dn.net/?f=y%3D-2.18%2C-0.22)
Answer:
43 1/12
Step-by-step explanation:
Add them together
First we have to have equal denominators
6 3/4 can also be 6 9/12
36 1/3 can also be 36 4/12
We add them to get 42 13/12
12/12 is 1 so we add 1 to 42 and get 43 1/12
Hope this helps
3+memes=frank123......................................