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ra1l [238]
3 years ago
6

What expression is equivalent to (^3√2^5)^1/4

Mathematics
2 answers:
Nezavi [6.7K]3 years ago
6 0
Soo first apply the exponent rule (2^5)^1/3*1/4
1/3*1/4= 1/12
so (2^5)^1/12
then you apply the exponent rule again 2^5*1/2
so 5*1/12= 5/12
So the answer would be 2^5/12
Helen [10]3 years ago
4 0
\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
\left( \sqrt[3]{2^5} \right)^{\frac{1}{4}}\implies \left( 2^{\frac{5}{3}} \right)^{\frac{1}{4}}\implies 2^{\frac{5}{3}\cdot \frac{1}{4}}\implies 2^{\frac{5}{12}}
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In an 80/20 mortgage, what is the second mortgage used for?
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Answer:

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Step-by-step explanation:

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3 years ago
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Two math students were asked to write an exponential growth equation that had a starting value of 300 and a growth rate of 2%. P
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Answer:

Pierre is right

Step-by-step explanation:

The correct formula for Exponential growth rate is given as:

y = a( 1 + r) ^t

Where

y = Amount after time t

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t = time

From the question

a = 300

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2 years ago
The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the
jolli1 [7]
If you've started pre-calculus, then you know that the derivative of  h(t)
is zero where h(t)  is maximum.

The derivative is            h'(t) = -32 t  +  96 .

At the maximum ...        h'(t) = 0

                                       32 t = 96 sec

                                           t  =  3 sec . 
___________________________________________

If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.

In that case, the question GIVES you the maximum height.
Just write it in place of  h(t), then solve the quadratic equation
and find out what  't'  must be at that height.

                                       150 ft = -16 t²  +  96  t  +  6 

Subtract 150ft from each side:    -16t²  +  96t  -  144  =  0 .

Before you attack that, you can divide each side by  -16,
making it a lot easier to handle:

                                                         t²  -  6t  +  9  =  0

I'm sure you can run with that equation now and solve it.    
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.  
(Funny how the two widely different methods lead to the same answer.)

The answer is from AL2006

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