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Luden [163]
3 years ago
9

3]{7x^6})^5" align="absmiddle" class="latex-formula">
How would I rewrite this using rational exponents? I know the answer is 7^\frac{5}{3}x^{10} , but I don't know how to get the x^{10}.
Mathematics
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

When you cube root x to the power of 6, x becomes squared, and the exponent, 5, makes it x to the power of 10.

Step-by-step explanation:

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Tim is buying packages of paper plates & cups. He wants to have equal amounts of each, but plates come in 24 per pack and cu
labwork [276]
D) 48

2 packs of plates is 48 plates and 3 packs of cups os 48 plates.
5 0
3 years ago
Which are sums of perfect cubes? Check all that apply.
vichka [17]
I think these are the sums of perfect cubes. 

A = (2x²)³ + (3)³
B = (x³)³ + (1)³
D = (x²)³ + (x)³
E = (3x³)³ + (x^4)³

5 0
3 years ago
If i have 18 apples and i get 1328 more how many apples would i have
devlian [24]

Darling, this problem is quite simply addition. You add 18 to 1328. 1328 + 18. we add the eights first. 8 + 8 equals 16. The 6 stays and we carry the one. 1 + 2 is 3. And then we add our carried 1 to make 4. So, the answer is 1346. I hope this helped!

6 0
4 years ago
Read 2 more answers
Radioactive Decay:
Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

<u>Step-by-step explanation:</u>

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}

The formula used to calculate the time period for a first order reaction follows:

t=\frac{2.303}{k}\log \frac{a}{(a-x)}

where,

k = rate constant = 0.01506days^{-1}

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

7 0
3 years ago
An aquarium has 12 equal size tanks of fish on display. All together, the tanks hold 643.8 gallons of water. How much water can
fgiga [73]

53.65 gallons → B

Since the tanks are equal in size then to find the volume of 1 tank

divide the total by 12

one tank = \frac{643.8}{12} = 53.65


6 0
3 years ago
Read 2 more answers
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