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3 years ago

How do i put (5x)-5/4 in radical form

1 answer:

5 0

Remove the negative exponent. $\frac{1}{(5x) \frac{5}{4} }$

If n is a positive integer that is greater than x and a is a real number or a factor, then $a \frac{x}{n} = \sqrt[n]{a ^{x} }$

Use the rule to convert (5x) $\frac{5}{4}$ to a radical, where a= 5x, x=5, and n=4.

so the answer would be..
$\frac{1}{ \sqrt[4]{(5x) ^{5} } }$

hope this helped and sorry if it was kind of confusing.

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What is the value of the expression 0.52?

polet [3.4K]

Answer:

0.25

Step-by-step explanation:

lets say you have half a cake

multiplying a decimal is the same as multiplying its counterpart

basically, you divide half a cake in half

0.25 of a cake, or 1/4

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3 years ago

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What value should go in the empty box to complete the calculation for finding the product of 62.834 × 0.45? A)2,213,360 B)2,313,

Aleonysh [2.5K]

Answer:

28.27530

Step-by-step explanation:

Given the expression 62.834 × 0.45, to solve this expression, first we need to convert it to fraction

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Take the product if the resulting fraction:

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3 years ago

One of the fastest recorded speeds of a cheetah is 26.8 m/s. Approximately, what is this speed in kilometers per hour? [1 kilome

ra1l [238]

We know that 1 kilometer = 1,000 meters and that 1 hour = 60 minutes = 3,600 seconds

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3 years ago

A dog chases a squirrel. The dog is originally 200 feet away from the squirrel. The dog's speed is 150 feet per minute. The squi

Dafna11 [192]

Answer:

15 seconds

Step-by-step explanation:

If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy: the dog's rate is 150 and the squirrel's is 100. The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know: the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up. Since we don't know what d is, we will just call it d! Now it seems as though we have 2 unknowns which is a problem. However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other. Here's the dog's equation:

d = rt

d+200 = 150t

And the squirrel's:

d = 100t

If we solve both for t and set them equal to each other we have:

$\frac{150}{d+200} =\frac{100}{d}$

Now we can cross multiply to solve for d:

150d = 100d + 20,000 and

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But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took. So sub that d value back into one of the equations we have solved for t and do the math:

$t=\frac{100}{d}=\frac{100}{400} =\frac{1}{4}$

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Can some one check this?

timama [110]

There are iInfinitely many solutions

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