![\displaysyle (\sqrt 2)^{3}=(2^{\frac{1}{2}})^3=2^{\frac{1}{2} \cdot 3}=2^{\frac{3}{2}}=\sqrt {2^3}=\sqrt 8](https://tex.z-dn.net/?f=%20%5Cdisplaysyle%20%28%5Csqrt%202%29%5E%7B3%7D%3D%282%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%29%5E3%3D2%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%203%7D%3D2%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%3D%5Csqrt%20%7B2%5E3%7D%3D%5Csqrt%208%20)
√8 is between 2 and 3, because 2²=4<8, but 3²=9>8. Also, our value is closer to 3 than to 2, so it is more than 2.5 and we have C and D options left.
Among these two numbers we find the one which is closer to √8.
C. 27=√729 ⇒ 2.7=√7.29
D. 28=√784 ⇒ <u>2.8=√7.84</u>
Hence our answer is D) 2.8
Answer:
1 2/3
Step-by-step explanation:
Answer:
A.
![y - 5 = -2(x-6)](https://tex.z-dn.net/?f=y%20-%205%20%3D%20-2%28x-6%29)
Negative reciprocal gives you the perpendicular slope so negative reciprocal of 1/2 is -2.
Then apply point-slope form.
B. The answer is x = 6.
The midpoint of JK is
![\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{8+ 4}{2}, \frac{4 + 4}{2} \right) = \left(6,4\right)](https://tex.z-dn.net/?f=%5Cleft%28%20%5Cfrac%7Bx_1%20%2B%20x_2%7D%7B2%7D%2C%20%5Cfrac%7By_1%20%2B%20y_2%7D%7B2%7D%20%5Cright%29%20%3D%20%5Cleft%28%20%5Cfrac%7B8%2B%204%7D%7B2%7D%2C%20%5Cfrac%7B4%20%2B%204%7D%7B2%7D%20%5Cright%29%20%3D%20%5Cleft%286%2C4%5Cright%29)
The line that goes through JK is just a horizontal line
because the y-coordinate does not change. So its perpendicular bisector is the vertical line that goes through the x-coordinate of the midpoint, that is,
.