The GCF of 8 and 6 is two
Answer: (y-c)/m
Reasoning:
Subtract c from both sides [ y-c=mx+c-c ]
Divide m from both sides [ (y-c)/m=mx/m ]
Answer:
Step-by-step explanation:
We know:
We have
Use 
![\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2%2B%5Ccos%5E2%5Ctheta%3D1%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B4%7D%2B%5Ccos%5E2%5Ctheta%3D1%5Cqquad%5Ctext%7Bsubtract%7D%5C%20%5Cdfrac%7B1%7D%7B4%7D%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5C%5Ccos%5E2%5Ctheta%3D%5Cdfrac%7B4%7D%7B4%7D-%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5C%5C%5Ccos%5E2%5Ctheta%3D%5Cdfrac%7B3%7D%7B4%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Csqrt%7B%5Cdfrac%7B3%7D%7B4%7D%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Cdfrac%7B%5Csqrt3%7D%7B%5Csqrt4%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Cdfrac%7B%5Csqrt3%7D%7B2%7D%5C%5C%5C%5C%5Ctheta%5Cin%5B0%5Eo%2C%5C%2090%5Eo%5D%2C%5C%20%5Ctext%7Btherefore%20all%20functions%20have%20positive%20values%20or%20equal%200.%7D%5C%5C%5C%5C%5Ccos%5Ctheta%3D%5Cdfrac%7B%5Csqrt3%7D%7B2%7D)
Answer:
a = 5
Step-by-step explanation:
Since the triangle is right use the sine ratio to solve for a and
sin30° = 
, thus
sin30° = 
= 
, hence
= ( cross- multiply )
2a = 10 ( divide both sides by 2 )
a = 5
Answer:
(-3, 3/2)
Step-by-step explanation:
To find the midpoint between two points you are going to add the X1 and X2, then divide by two. Then you are going to add the Y1 and Y2, and divide by two.
A(-6,1) B(0,2)
= (-6+0, 1+2)
= (-6, 3)
=(-6/2, 3/2)
=(-3, 3/2)
