<u>Solution </u><u>1</u><u> </u><u>:</u><u>-</u>
Given expression ,
We know that, ![a^{\frac{m}{n}} = \sqrt[n]{a^m}](https://tex.z-dn.net/?f=%20a%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%20%3D%20%5Csqrt%5Bn%5D%7Ba%5Em%7D%20)
Therefore , this can be written as ,
![\implies \sqrt[4]{13^3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B4%5D%7B13%5E3%7D%20)
<u>Solution</u><u> </u><u>2</u><u>:</u><u>-</u>
Given expression ,
![\implies \sqrt[3]{24^{10}}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B3%5D%7B24%5E%7B10%7D%7D%20)
We know that,
Therefore, this can be written as ,
Make a system of equations and solve for each variable.
3m+8v=53
5m+2v=26
m= movie cost
v= game cost
multiply the bottom equation by 4(combination method)
3m+8v=53
20m+8v=104. Subtract
-17m = -51
m= 3
Plug back into one of the equations
5(3)+2v=26
15+2v=26
2v=11
v=5.5
m=$3
v=$5.5
Answer:
c
Step-by-step explanation:
Answer:
11z/4+3/10
Step-by-step explanation:
Distribute first then combine like terms.
Your friend's distance from her starting point to your position is decreasing until she passes by you. This distance is D(t) = 100 ft - (20 ft / sec)t. We'll assume that your friend doesn't actually pass by you; she just gets closer and closer, from her initial 100 ft distance from you. Then:
D(t) = 40 ft = 100 ft - (20 ft / sec)t. Simplifying, we get:
-60 ft = -(20 ft / sec)t, and thus t = 3 sec.
