Answer:to find the difference between 6/8 and 1/7, you need to make the denominators (the bottom numbers) the same. turn 6/8 into 42/56. then turn 1/7 into 8/56. you’ll then do 42/56-8/56. so it’ll be 42-8=36. so it will come out to 36/56, which is (when simplified) is 9/14. your answer will be 9/14.
Step-by-step explanation:
Answer: The answer is given below
Step-by-step explanation:
From the question, we are informed that the money m (in £) Xavier has to spend each week is his wage w (in £) subtract the tax t (in £) he pays on his income. Therefore, the formula for the money he can spend will be:
m = w - t
The amount of money that he has to spend if he earns £120 a week and pays £41 in tax will be:
m = w - t
m = £120 - £41
m = £79
Hi,
The answer is 8(6/9) = 6(8/9) as both are equal to 48/9 = 16/3
Hope this helps! If you’d like further explanation please let me know.
Answer:
The triangles ABE and MNP are similar because the three corresponding angles are equal (AAA Criteria)
Step-by-step explanation:
we know that
An isosceles triangle has two equal sides and two equal angles. The two equal angles are called the base angles and the third angle is called the vertex angle
The triangle ABE is an isosceles triangle with the vertex angle m∠ABE equal to
so
m∠BAE=m∠AEB-------> base angles
m∠BAE+m∠AEB+m∠ABE=
------> sum of internal angles of triangle
Find m∠BAE
2m∠BAE=
m∠BAE=
The measure of the angles of triangle ABE are 
The triangle MNP is an isosceles triangle with the base angle m∠NMP and m∠NPM equal to
so
m∠MNP+m∠NMP+m∠NPM=
------> sum of internal angles of triangle
Find m∠MNP (vertex angle)
m∠MNP=
m∠MNP=
The measure of the angles of triangle MNP are 
therefore
The triangles ABE and MNP are similar because the three corresponding angles are equal (AAA Criteria)
Answer:

Step-by-step explanation:
![\cos(2x) = \cos^2 x-\sin^2 x = 1-2\sin^2 x \\ \\ \cos(x) = 1-2\sin^2 (\frac{x}{2}) \\ \\ \Rightarrow \sin^2 (\frac{x}{2}) = \dfrac{1-\cos(x)}{2}\\ \\ \sin(\frac{x}{2}) = \pm \sqrt{\dfrac{1-\cos(x)}{2}},\quad x\in [\frac{3\pi }{2},\pi] \Rightarrow \frac{x}{2}\in [\frac{3\pi}{4},\frac{\pi}{2}]\\ \\ \Rightarrow \sin(\frac{x}{2}) > 0 \Rightarrow \sin(\frac{x}{2}) = \sqrt{\dfrac{1-(-\frac{3}{5})}{2}} \Rightarrow \sin(\frac{x}{2}) = \sqrt{\dfrac{8}{10}}=\dfrac{2\sqrt 2}{\sqrt{10}} = \\ \\ =\dfrac{2\sqrt 5}{5}](https://tex.z-dn.net/?f=%5Ccos%282x%29%20%3D%20%5Ccos%5E2%20x-%5Csin%5E2%20x%20%3D%201-2%5Csin%5E2%20x%20%5C%5C%20%5C%5C%20%5Ccos%28x%29%20%3D%201-2%5Csin%5E2%20%28%5Cfrac%7Bx%7D%7B2%7D%29%20%5C%5C%20%5C%5C%20%5CRightarrow%20%5Csin%5E2%20%28%5Cfrac%7Bx%7D%7B2%7D%29%20%3D%20%5Cdfrac%7B1-%5Ccos%28x%29%7D%7B2%7D%5C%5C%20%5C%5C%20%5Csin%28%5Cfrac%7Bx%7D%7B2%7D%29%20%3D%20%5Cpm%20%5Csqrt%7B%5Cdfrac%7B1-%5Ccos%28x%29%7D%7B2%7D%7D%2C%5Cquad%20x%5Cin%20%5B%5Cfrac%7B3%5Cpi%20%7D%7B2%7D%2C%5Cpi%5D%20%5CRightarrow%20%5Cfrac%7Bx%7D%7B2%7D%5Cin%20%5B%5Cfrac%7B3%5Cpi%7D%7B4%7D%2C%5Cfrac%7B%5Cpi%7D%7B2%7D%5D%5C%5C%20%5C%5C%20%5CRightarrow%20%5Csin%28%5Cfrac%7Bx%7D%7B2%7D%29%20%3E%200%20%5CRightarrow%20%5Csin%28%5Cfrac%7Bx%7D%7B2%7D%29%20%3D%20%5Csqrt%7B%5Cdfrac%7B1-%28-%5Cfrac%7B3%7D%7B5%7D%29%7D%7B2%7D%7D%20%5CRightarrow%20%5Csin%28%5Cfrac%7Bx%7D%7B2%7D%29%20%3D%20%5Csqrt%7B%5Cdfrac%7B8%7D%7B10%7D%7D%3D%5Cdfrac%7B2%5Csqrt%202%7D%7B%5Csqrt%7B10%7D%7D%20%3D%20%5C%5C%20%5C%5C%20%3D%5Cdfrac%7B2%5Csqrt%205%7D%7B5%7D)