Answer:yes
Step-by-step explanation:
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Since we know that the hypotenuse is 11 and this is a 30 - 60 - 90 right triangle we can use 11 to find the other sides.
The hypotenuse is 2s, s is the 30 side length.
So s = 11 / 2 = 5.5
The 60 side is s * square root of 3.
5.5 * square root of 3 = about 9.5.
Now that we know all three sides we can add them together to get the perimeter.
11 + 5.5 + 9.5 = 26
:)))
Substitute y = 1/5 x + 1 into x + 5y = -15
x + 5y = -15
x + 5(1/5 x + 1) = -15
x + x + 5 = -15
2x = -20
x = -10
Plug in x = -10 into y = 1/5 x + 1
y = 1/5 (-10) + 1
y = -2 + 1
y = -1
(-10 , -1)
Answer is the last one
(-10 , -1)
Your statement that angle C and angle D adding to 180 degrees isn't always true. What is always true is that angle A and angle C add to 180 degrees.
Rule: The opposite angles of an inscribed quadrilateral are supplementary (add to 180 degrees)
So,
(angle A) + (angle C) = 180
(x+2) + (x-2) = 180
x+2+x-2 = 180
2x = 180
2x/2 = 180/2
x = 90
Once you know x, you can find the angles
measure of angle A = x+2
measure of angle A = 90+2
measure of angle A = 92 degrees
measure of angle C = x-2
measure of angle C = 90-2
measure of angle C = 88 degrees
Note how A+C = 92+88 = 180
measure of angle D = x-10
measure of angle D = 90-10
measure of angle D = 80 degrees
measure of angle B = 360-A-C-D ... see note below
measure of angle B = 360-92-88-80
measure of angle B = 100 degrees
Note: for any quadrilateral, the four angles add to 360 degrees. So A+B+C+D = 360. We can solve for B to get B = 360-A-C-D
