Answer:
=![\sqrt[3]{49x^{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B49x%5E%7B2%7D%20%7D)
Step-by-step explanation:
We have been given the expression;
7 times x to the two thirds power which can be written mathematically as;
To express the above expression as a radical, we need to recall that;
![a^{\frac{b}{n}}=\sqrt[n]{a^{b}}](https://tex.z-dn.net/?f=a%5E%7B%5Cfrac%7Bb%7D%7Bn%7D%7D%3D%5Csqrt%5Bn%5D%7Ba%5E%7Bb%7D%7D)
Therefore;
![(7x)^{\frac{2}{3}}=\sqrt[3]{(7x)^{2} }\\\\=\sqrt[3]{49x^{2} }](https://tex.z-dn.net/?f=%287x%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7B%287x%29%5E%7B2%7D%20%7D%5C%5C%5C%5C%3D%5Csqrt%5B3%5D%7B49x%5E%7B2%7D%20%7D)
It’s gonna be 7(8+3) you use distributive property so it’ll be the same as 56+21
Answer:
We get that the 165.1% of 333 is 333 • 1.651 = 549.78
Step-by-step explanation:
1. They are asking for the 165.41%
2. So, we need to multiply 333 by the representation of 165.1% in decimal by doing the multiplication using a calculator
Answer:
The number of liters of 50% antifreeze solution is 50 and the number of liters of 90% antifreeze solution is 150
Step-by-step explanation:
Let
x ----> number of liters of 50% antifreeze solution
y ----> number of liters of 90% antifreeze solution
we know that
50%=50/100=0.50
90%=90/100=0.90
80%=80/100=0.80
x+y=200
x=200-y ------> equation A
0.50x+0.90y=0.8(200) -----> equation B
substitute equation A in equation B and solve for y
0.50(200-y)+0.90y=160
100-0.50y+0.90y=160
0.40y=160-100
0.40y=60
y=150 liters
Find the value of x
x=200-y
x=200-150=50 liters
therefore
The number of liters of 50% antifreeze solution is 50
The number of liters of 90% antifreeze solution is 150
Hey!
Switch it around and it becomes:
First find 18 ÷ -6
Positive ÷ negative OR negative ÷ positive is always negative.
That leaves you with:
Switch that around and it becomes:
Find -3 + 5
That leaves you with:
Divide both sides by 2 to leave <em>x</em> alone
