36 there you go I hope you understand
Answer:
x = 16
y = 20
Step-by-step explanation:
the left part of the triangle can be solved first as it already has one angle and 1 side, we use trigonometry for this.
since we have 2 sides and 1 angle, to find the third side length we use the pythagoras theorem. a² + b² = c²
; 9² + 12² = c²
; 225 = c²
;√225 = c
; 15 = c
since we have all the lengths of the triangle, we can find out the angle opposite to the side length '9',
;
;
x 9 = Sinx
36.86 = x
now, we look at the triangle as a whole, the top angle being 90 degrees, we know that we have 36.86.
; 90 - 36.86 = 53.13

;
x Sin 90 = 20
y = 20
again, since we have 2 sides of a right angled triangle we use a² + b² = c²
we arrange this formula since we have the hypotenuse
; 20² - 12² = 256
; √256 = x
; 16 = x
473 x 19 = 8987. Your welcome and i no speak espanol
<u>Answer:
</u>
The point slope form of the line that passes through (-8,2) and is parallel to a line with a slope of -8 is 8x + y + 62 = 0
<u>Solution:
</u>
The point slope form of the line that passes through the points
and parallel to the line with slope “m” is given as
--- eqn 1
Where “m” is the slope of the line.
are the points that passes through the line.
From question, given that slope “m” = -8
Given that the line passes through the points (-8,2).Hence we get

By substituting the values in eqn 1, we get the point slope form of the line which is parallel to the line having slope -8 can be found out.
y-2=-8(x-(-8))
On simplifying we get
y – 2 = -8(x +8)
y – 2 = -8x -64
y – 2 +8x +64 = 0
8x + y +62 = 0
Hence the point slope form of given line is 8x + y +62 = 0
Problem 1 is fully factored as each term is a binomial raised to some exponent. If the exponent isn't showing up, it's because it is 1. Recall that x^1 = x.
Problem 2 can be factored further because x^2-8x+16 factors to (x-4)(x-4) or (x-4)^2. To get this factorization, you find two numbers that multiply to 16 and add to -8. Those two numbers are -4 and -4 which is where the (x-4)(x-4) comes from. Overall, the entire thing factors to (x-4)^2*(x+3)*(x-2)
Problem 3 is a similar story. We can factor x^2-1 into (x-1)(x+1). I used the difference of squares rule here. Or you can think of x^2-1 as x^2+0x-1, then find two numbers that multiply to -1 and add to 0. Those two numbers are +1 and -1 which leads to (x+1)(x-1). So the full factorization is (x-1)(x+1)(x+1)(x-4) which is the same as (x-1)(x+1)^2(x-4)