Answer:
A. $1,110
B. 50 units.
Step-by-step explanation:
A. Substitute
into the function
for calculate the time in hours required for the worker to produce 18 units:
![T(18)=4(18)+2=74](https://tex.z-dn.net/?f=T%2818%29%3D4%2818%29%2B2%3D74)
Now you need to substitute
into the function
to calculate the amount in dollars she will be paid for 74 hours:
![S(74)=15(74)=1,110](https://tex.z-dn.net/?f=S%2874%29%3D15%2874%29%3D1%2C110)
She will be paid $1,110.
B. To calculate the number of units that the worker needs to produce to be paid $3,030 , you need to substitute
into the function
and solve for T (to find the number of hours she requires to work to be paid $3,030)
![3,030=15T\\\\T=\frac{3,030}{15}\\\\T=202](https://tex.z-dn.net/?f=3%2C030%3D15T%5C%5C%5C%5CT%3D%5Cfrac%7B3%2C030%7D%7B15%7D%5C%5C%5C%5CT%3D202)
Now, substitute
into
and solve for x, to calculate the number of units she produces in 202 hours:
![202=4x+2\\\\202-2=4x\\\\x=\frac{200}{4}\\\\x=50](https://tex.z-dn.net/?f=202%3D4x%2B2%5C%5C%5C%5C202-2%3D4x%5C%5C%5C%5Cx%3D%5Cfrac%7B200%7D%7B4%7D%5C%5C%5C%5Cx%3D50)
She would need to produce 50 units to be paid $3,030
Answer: B (3 cups)
Step-by-step explanation:
Add 2 cups + 1 cup together, and you get 3 cups. C is incorrect (in this scenario) because it doesn’t describe what measurement it’s in.
Answer:
1- 5xy³√5y
2- 2xy²∛3y²
Step-by-step explanation:
√125x²y^7=
√25*5x²y^6y
5xy³√5y
2) ∛24x³y^8=
∛2³*3x³y^8=
2xy²∛3y²
The correct answer is 4 ( 6x - 1 )
We'll use the following properties of sine and cosine to prove this:
![sin(a)\cdot sin(b) = \frac12(cos(a-b) - cos(a+b))](https://tex.z-dn.net/?f=sin%28a%29%5Ccdot%20sin%28b%29%20%3D%20%5Cfrac12%28cos%28a-b%29%20-%20cos%28a%2Bb%29%29)
![cos(a)\cdot cos(b) = \frac12(cos(a+b) + cos(a-b))](https://tex.z-dn.net/?f=cos%28a%29%5Ccdot%20cos%28b%29%20%3D%20%5Cfrac12%28cos%28a%2Bb%29%20%2B%20cos%28a-b%29%29)
![cos(180+a) = cos(180-a)](https://tex.z-dn.net/?f=cos%28180%2Ba%29%20%3D%20cos%28180-a%29)
Then it's just a matter of filling it in...
sin20sin40 * sin60sin80 = 1/2(cos20 - cos60) * 1/2 (cos20 - cos140) =
1/8( cos40 + 1 - cos160 - cos120 - cos40 - cos80 + cos80 + cos200) =
1/8(1 - cos160 - cos120 + cos200) =
1/8(1 - cos160 - cos120 + cos160) =
1/8(1 - cos120 ) = 1/8( 1 + 1/2 ) = 3/16