1) c- you have to imagine if you folded it and have b and f where they are pictured on the right
2) a- supplementary angles add up to 180°
Answer:
x=0
y=3
Step-by-step explanation:
y =7x+3 equation 1
7x-6y=-18 equation 2
using equation 1 in equation 2 we have:
7x-6(7x+3)= -18
7x -42x -18 = -18
-35x -18 = -18
-35x = -18 +18
-35x=0
x=0
so we have:
y =7x+3
y= 7(0) +3
y = 3
Answer:
![x^2+12x-5=(x+6)^2-41](https://tex.z-dn.net/?f=x%5E2%2B12x-5%3D%28x%2B6%29%5E2-41)
Step-by-step explanation:
The given expression is ![x^2+12x-5](https://tex.z-dn.net/?f=x%5E2%2B12x-5)
Comparing to
, we have ![a=1,b=12,c=-5](https://tex.z-dn.net/?f=a%3D1%2Cb%3D12%2Cc%3D-5)
We add and subtract ![(\frac{b}{2a})^2](https://tex.z-dn.net/?f=%28%5Cfrac%7Bb%7D%7B2a%7D%29%5E2)
![x^2+12x+6^2-6^2-5](https://tex.z-dn.net/?f=x%5E2%2B12x%2B6%5E2-6%5E2-5)
This implies that:
![\boxed{x^2+6x+36}-36-5](https://tex.z-dn.net/?f=%5Cboxed%7Bx%5E2%2B6x%2B36%7D-36-5)
The expression in the rectangle is a perfect square
![(x+6)^2-41](https://tex.z-dn.net/?f=%28x%2B6%29%5E2-41)
Therefore ![x^2+12x-5=(x+6)^2-41](https://tex.z-dn.net/?f=x%5E2%2B12x-5%3D%28x%2B6%29%5E2-41)
The trigonometry ratios are cos(θ) = -7/√58, sec(θ) = -√58/7 and cot(θ) = -7/3
<h3>How to determine the trigonometry ratios?</h3>
The point on the terminal side is given as:
(x, y) = (-7, 3)
Start by calculating the hypotenuse using:
![h= \sqrt{x^2 + y^2](https://tex.z-dn.net/?f=h%3D%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2)
So, we have:
![h= \sqrt{(-7)^2 + 3^2](https://tex.z-dn.net/?f=h%3D%20%5Csqrt%7B%28-7%29%5E2%20%2B%203%5E2)
Evaluate
![h= \sqrt{58](https://tex.z-dn.net/?f=h%3D%20%5Csqrt%7B58)
The cosine is then calculated using:
cos(θ) = x/h
This gives
cos(θ) = -7/√58
The secant is then calculated using:
sec(θ) = 1/cos(θ)
This gives
sec(θ) = -√58/7
The cotangent is then calculated using:
cot(θ) = cos(θ)/sin(θ)
Where
sin(θ) = y/h
So, we have:
sin(θ) = 3/√58
So, we have:
cot(θ) = (-7/√58)/(3/√58)
This gives
cot(θ) = -7/3
Read more about terminal points at:
brainly.com/question/1621860
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