There's more than one way to combine them really
but an obvious one will be
![\bf \begin{array}{llll} h(x)&=&(f\circ g)(x)\\\\ &&\sqrt[3]{7x+1}\\\\ &&\sqrt[3]{g(x)}\leftarrow \begin{array}{llll} f(x)=\sqrt[3]{x }\\\\ g(x)=7x+1 \end{array} \end{array}\\\\ -----------------------------\\\\ (f\circ g)(x)\iff f[\quad g(x)\quad ]=\sqrt[3]{g(x)}\implies f[\quad g(x)\quad ]=\sqrt[3]{7x+1}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0Ah%28x%29%26%3D%26%28f%5Ccirc%20g%29%28x%29%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7B7x%2B1%7D%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cleftarrow%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Af%28x%29%3D%5Csqrt%5B3%5D%7Bx%20%7D%5C%5C%5C%5C%0Ag%28x%29%3D7x%2B1%0A%5Cend%7Barray%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%28f%5Ccirc%20g%29%28x%29%5Ciff%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cimplies%20%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7B7x%2B1%7D)
She owes in total 40$ in total but seprate
Step-by-step explanation:
y = ax + b
I see already the result (y = x/10 × 1/6), but let's go in formally.
we have multiple function points to use to officially calculate a and b.
1/6 = a×10 + b
2/3 = a×40 + b
5/6 = a×50 + b
1 2/3 = a×100 + b
let's e.g subtract equation 1 from equation 3.
4/6 = a×40 + 0
a = 4/40 / 6 = 4 / 240 = 1/60
1/6 = 10/60 + b
1/6 = 1/6 + b
b = 0
so, the function is
y = x/60
x = 25
y = 25/60 = 5/12 cups
For the equation to work, you have to have the same number on both sides of the equal sign
so to find ?, you can do:
13 - 2 = 12 - ? Simplify
11 = 12 - ? Subtract 12 on both sides
-1 = - ? Multiply -1 on both sides
1 = ?
Answer:
m∠PNO = 60°
m∠O = 33°
Step-by-step explanation:
∡NPO is 87° because it's a vertical angle with the 87° angle
3) ∡PNO is supplementary to the 120° angle (they must add to 180°)
4) m∠O = 180 - (60 + 87) = 33
