Answer:
Step-by-step explanation:
y = -5/4x + 5
Answer:
A
Step-by-step explanation:
sin θ=3/8
θ=sin ^{-1}(3/8)
≈22.02 °
The cable should be 50 feet.
If you draw the picture, you will see that you have a right triangle.
Let's just use the Pythagorean Theorem to find the hypotenuse (or cable).
30^2 + 40^2 = c^2
2500 = c^2
50 = c
Answer:
A
Step-by-step explanation:
if (x1,y1) and (x2,y2) are the extremities of diameter,then eq. of circle is
(x-x1)(x-x2)+(y-y1)(y-y2)=0
reqd. eq. is (x+1)(x-5)+(y+9)(y-1)=0

center is (2,-4)
r=√(2²+(-4)²-(-14))
=√(4+16+14)
=√(34)
eq. of circle is (x-2)²+(y+4)²=34
or
(x²-4x)+(y²+8y)=14
(x²-4x+4)+(y²+8y+16)=14+4+16
(x-2)²+(y+4)²=34
Given: The following functions



To Determine: The trigonometry identities given in the functions
Solution
Verify each of the given function

B

C

D

E

Hence, the following are identities

The marked are the trigonometric identities