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erastova [34]
2 years ago
15

Help someone how do you solve this

Mathematics
1 answer:
Flauer [41]2 years ago
5 0

Answer:

a = 10/3

Step-by-step explanation:

This can be written as:

x^{2} (x^{4})^{1/3}  = x^{a}

Anything to the power of 1/3 means to find the cube root. We can simplify this to:

x^{2} (x^{4/3}) = x^{a}

I just multiplied 4 and 1/3 (because of the second index law (a^{m})^{n}  = a^{mn}). Now, we can add the powers because the bases are the same and we are multiplying (first index law):

x^{10/3} = x^{a}

a = 10/3

Hope this helps!

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Answer: imagine not knowing the answer

Step-by-step explanation:

5 0
2 years ago
The graph shown is a translation of the graph of ​f(x)=x^2. Write the function in vertex form.
Gnesinka [82]

\sf \longrightarrow \: f(x) =  {x}^{2}

Write the parent function in the standard form I.e.

\boxed{ \tt \:y =a(x - h) \times 2 + k  }

Where,

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\sf \longrightarrow \:y=  1(x - 0)^{2} \times 2 + 0

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4 0
2 years ago
Help please<br>Mark as brainliest
34kurt
You need to use this formula:
([a]/[sinA])=([c]/[sinC])-I am going to use 'a' for the x, and 'c' for 16(square root of 3)
Now its just getting 'a' by itself.
([c] times [sinA])/([sinC])=[a]
[c]=16 square root of 3
sinA=sin(60)=(square root of 3)/2
sinC=sin(90)=1

Plug it in to get 24 for a, or x. Do the same to figure out y with the new sides.
The final y is:
8 square root of 3
Final answer is C.
8 0
3 years ago
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If they are, the perceptive is a bit different. Your expression gives the likelihood that a particular set of j balls goes into the last urn and the other n−j balls into the other urns. But there are (nj) different possible sets of j balls, and each of them the same probability of being the last insides of the last urn, so the total probability of completing up with exactly j balls in the last urn is if the balls are different.


See attached file for the answer.

3 0
3 years ago
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