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Answer:</h2>
To solve these, we will divide on both.
![5y = 120\\\\y = 24](https://tex.z-dn.net/?f=5y%20%3D%20120%5C%5C%5C%5Cy%20%3D%2024)
![2x = 108\\\\x = 54](https://tex.z-dn.net/?f=2x%20%3D%20108%5C%5C%5C%5Cx%20%3D%2054)
Answer:
A
Step-by-step explanation:
We are given:
![\displaystyle \cos(\theta)=-\frac{12}{13},\,\theta\in\text{QIII}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccos%28%5Ctheta%29%3D-%5Cfrac%7B12%7D%7B13%7D%2C%5C%2C%5Ctheta%5Cin%5Ctext%7BQIII%7D)
Since cosine is the ratio of the adjacent side over the hypotenuse, this means that the opposite side is (we can ignore negatives for now):
![o=\sqrt{13^2-12^2}=\sqrt{25}=5](https://tex.z-dn.net/?f=o%3D%5Csqrt%7B13%5E2-12%5E2%7D%3D%5Csqrt%7B25%7D%3D5)
So, the opposite side is 5, the adjacent side is 12, and the hypotenuse is 13.
And since θ is in QIII, sine/cosecant is negative, cosine/secant is negative, and tangent/cotangent is positive.
Cosecant is given by the hypotenuse over the opposite side. Thus:
![\displaystyle \csc(\theta)=\frac{13}{5}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccsc%28%5Ctheta%29%3D%5Cfrac%7B13%7D%7B5%7D)
Since θ is in QIII, cosecant must be negative:
![\displaystyle \csc(\theta)=-\frac{13}{5}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccsc%28%5Ctheta%29%3D-%5Cfrac%7B13%7D%7B5%7D)
Our answer is A.
F(x) = x^2 - 6x + 4 {-3, 0, 5} (x, y) x is the domain, y is the range.
plug in the domain numbers into the equation to find the range.
y = x^2 - 6x + 4
y = -3^2 - 6(-3) + 4
y = 9 - (-18) + 4
y = 9 + 18 + 4
y = 31 (-3, 31)
y = x^2 - 6x + 4
y = 0^2 - 6(0) + 4
y = 0 - 0 + 4
y = 4 (0, 4)
y = x^2 - 6x + 4
y = 5^2 - 6(5) + 4
y = 25 - 30 + 4
y = -1 (5, -1)
c. {-1, 4, 31}
hope this helped, God bless!
Y = sallys grocerys
x = how much sally paid
y - (.05) = x
Answer:
112 if you don't include the triangle sides as they are not necessary, but 124 if you include one triangle and 136 if you include both triangle sides