Opposite angles in a cyclic quadrilateral add up to 180 degrees .
So the angle opposite angle A ( that is angle C) must be 180-95 = 85 degrees
Answer:

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel
And the anfle is approximately 
Step-by-step explanation:
For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.
a=[1,2,-2], b=[4,0,-3,]
The dot product on this case is:

Since the dot product is not equal to zero then the two vectors are not orthogonal.
Now we can calculate the magnitude of each vector like this:


And finally we can calculate the angle between the vectors like this:

And the angle is given by:

If we replace we got:

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel
And the anfle is approximately 
5 2/5 would be converted to 5 6/15. you would then subtract the numerators and the whole numbers, giving you 3 5/15, which simplifies to 3 1/5
9514 1404 393
Answer:
x = 8/7
Step-by-step explanation:
.2(.7x-5)+0.2=1.4(x-1.6) . . . . given
Eliminating parentheses, we have ...
0.14x -1 +0.2 = 1.4x -2.24
1.44 = 1.26x . . . . . . . add 2.24-0.14x to both sides
x = 1.44/1.26 . . . . . . divide by the coefficient of x
x = 8/7 . . . . . . . . . . .simplify the fraction
__
As a repeating decimal, the value of x is ...

However, the notation 1.(142857) is usually reserved for the case where the contents of parentheses are being multiplied by the factor in front. You would need to consult your instructions to see how the decimal version of the answer should be entered. (Quite often, rounding is suggested.)
_____
<em>Additional comment</em>
The equation in the text of your question is different from the one in the picture. We have answered for the one in the picture. The text asks for the solution of a quadratic. Those two solutions are approximately 1.085 and 16.057.
If you're using the app, try seeing this answer through your browser: brainly.com/question/2822258_______________
• Function: f(x) = 3x + 12.
A. Finding the inverse of f.
The composition of f with its inverse results in the identity function:
(f o g)(x) = x
f[ g(x) ] = x
3 · g(x) + 12 = x
3 · g(x) = x – 12
x – 12
g(x) = ⸺⸺
3
x g(x) = ⸺ – 4 <——— this is the inverse of f.
3________
B. Verifying that the composition of f and g gives us the identity function:
•

![\mathsf{=f\big[g(x)\big]}\\\\\\ \mathsf{=3\cdot \left(\dfrac{x}{3}-4\right)+12}\\\\\\ \mathsf{=\diagup\hspace{-7}3\cdot \dfrac{x}{\diagup\hspace{-7}3}-3\cdot 4+12}\\\\\\ \mathsf{=x-12+12}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Df%5Cbig%5Bg%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D3%5Ccdot%20%5Cleft%28%5Cdfrac%7Bx%7D%7B3%7D-4%5Cright%29%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%5Cdfrac%7Bx%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-3%5Ccdot%204%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx-12%2B12%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
and also
•

![\mathsf{=g\big[f(x)\big]}\\\\\\ \mathsf{=\dfrac{f(x)}{3}-4}\\\\\\ \mathsf{=\dfrac{3x+12}{3}-4}\\\\\\ \mathsf{=\dfrac{\diagup\hspace{-7}3\cdot (x+4)}{\diagup\hspace{-7}3}-4}\\\\\\ \mathsf{=x+4-4}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Dg%5Cbig%5Bf%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7Bf%28x%29%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7B3x%2B12%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdfrac%7B%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%28x%2B4%29%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%2B4-4%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
________
C. Since f and g are inverse, then
f(g(– 2))
= (f o g)(– 2)
=
– 2 <span>✔
</span>
• Call h the compositon of f and g. So,
h(x) = (f o g)(x)
h(x) = x
As you can see above, there is no restriction for h. Therefore, the domain of h is R (all real numbers).
I hope this helps. =)