Answer:
7()=11
8()=42
I could barely see the answer for number 9 sorry :(
Step-by-step explanation:
the 7() is 2(3)+5
2(3)=6
6+5=11
Question (8)
c^2+3a x b
c(6)^2+ 3(3) x 2/3
36+9 x 2/3
36+6=42
Answer:
C
Step-by-step explanation:
move the x over
-x+y=6
move the y over
-x = -y+6
times each number by a -
x=y-6
Answer:
16
Step-by-step explanation:
First, substitute the values of the variables into the expression:
3(y + 4) - (x - 1)
3(2 + 4) - (3 - 1)
Now, simplify both parentheses:
3(6) - 2
18 - 2
Finally, simplify:
16
Answer:
option c is correct.
Step-by-step explanation:
![7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)](https://tex.z-dn.net/?f=7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B16x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B8x%7D%5Cright%29)
WE need to simplify this equation.
Solve the parenthesis of each term.
![=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right](https://tex.z-dn.net/?f=%3D7%5Cleft%5Csqrt%5B3%5D%7B2x%7D%5Cright-3%5Cleft%5Csqrt%5B3%5D%7B16x%7D%5Cright-3%5Cleft%5Csqrt%5B3%5D%7B8x%7D%5Cright)
Now, We will find factors of the terms inside the square root
factors of 2: 2
factors of 16 : 2x2x2x2
factors of 8: 2x2x2
Putting these values in our equation:![=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)](https://tex.z-dn.net/?f=%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2X2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%20x%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%7D%20%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%7D%20%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2%5E3%7D%20%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2%5E3%7D%20%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%2A2%5Cleft%28%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%2A2%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%5Cright%29-6%5Cleft%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29)
Adding like terms we get:
![=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\](https://tex.z-dn.net/?f=%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%5Cright%29-6%5Cleft%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%5C%5C%3D%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C)
![(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\can\,\,be \,\, written\,\, as\,\,\\(\sqrt[3] {2x})-6\left(\sqrt[3]{x}\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5Ccan%5C%2C%5C%2Cbe%20%5C%2C%5C%2C%20written%5C%2C%5C%2C%20as%5C%2C%5C%2C%5C%5C%28%5Csqrt%5B3%5D%20%7B2x%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29)
So, option c is correct