8.96 * 30%
30 ->.3
8.96 * .3 = 2.69
8.96 + 2.69 = 11.65
![x^ \frac{m}{n}= \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7Bm%7D%7Bn%7D%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%20)
pemdas, so exponent first before multiply
4(x^1/2)=4x^2
this is different from
(4x)^1/2
so
![x^ \frac{1}{2}= \sqrt[2]{x^1}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7B1%7D%7B2%7D%3D%20%5Csqrt%5B2%5D%7Bx%5E1%7D%20%20)
times that by 4
4√x
Step-by-step explanation:
Let grandson's age be x.
Let grandpa's age be y {equation : y = 8+5x}
so,
x + y = 88
or, x + 8+ 5x = 88 (substituting the value of y)
or, 6x + 8 =88
or, 6x = 88-8
or, 6x = 80
or, x = 80/6
so, x = 13.33
so grandson's age = 13.33
grandpa's age = 8 + 5x = 8 + (5×13.33) = 74.65 yrs
The answer is D. You can't make an equation out of this!
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15% of 20 gal = 0.15 * 20 gal = 3 gal, so the solution contains 3 gal of salt.
If we add <em>x</em> gal of water to the solution, we end up with (20 + <em>x</em>) gal of solution. We want the new mixture to have a concentration of 10%, or
10% of (20 + <em>x</em>) gal = 0.1 * (20 + <em>x</em>) gal = 2 + 0.1<em>x</em> gal
of salt.
The amount of salt in the tank hasn't changed. Solve for <em>x </em>:
2 + 0.1<em>x</em> = 3
0.1<em>x</em> = 1
<em>x</em> = 10 gal