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Natali [406]
3 years ago
14

What is a expression equivalent to ^3√8x^2y^3z^4

Mathematics
1 answer:
Daniel [21]3 years ago
4 0
3sqrt(8x^2y^3z^4)
=2sqrt(8)xy^(3/2)z
=4sqrt(2)xzy^(3/2)
hope it helps :)
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If I put a different sticker on the 30th, 50th, and 60th pages will any of the first 600 pages have all 3 stickers?
icang [17]

Answer:

so the which page that will have the 3 sticker, we must solve the least common multiple of 30, 50, 60. A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24, ....

The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.

so the least common multiple of 30, 50 and 60 is 300. so the page that will have 3 stickers is 300th page

Step-by-step explanation:

8 0
3 years ago
How much more is 9,00,000 than 4,50,000
KIM [24]

Answer: 4,50,000 or 63

Step-by-step explanation: I add

5 0
3 years ago
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If you were creating a lamp by hand what type of material would you choose for the wiring in what material would you choose to c
grin007 [14]

The material for the lamp of the wiring would be cutting with pliers.

8 0
3 years ago
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The simple interest on a certain sum of money for 2 years at 5% per annum is Rs 320. What will be the compound interest on the s
Brilliant_brown [7]

Answer:

Rs 328

Step-by-step explanation:

Find the <u>principal</u> amount invested.

<u>Simple Interest Formula</u>

I = Prt

where:

  • I = interest earned
  • P = principal
  • r = interest rate (in decimal form)
  • t = time (in years)

Given:

  • I = Rs 320
  • r = 5% = 0.05
  • t = 2 years

Substitute the given values into the formula and solve for P:

⇒ 320 = P(0.05)(2)

⇒ 320 = P(0.1)

⇒ P = 3200

<u>Compound Interest Formula</u>

\large \text{$ \sf I=P\left(1+\frac{r}{n}\right)^{nt} -P$}

where:

  • I = interest earned
  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Given:

  • P = 3200
  • r = 5% = 0.05
  • n = 1 (annually)
  • t = 2 years

Substitute the given values into the formula and solve for I:

\implies \sf I=3200\left(1+\frac{0.05}{1}\right)^{2} -3200

\implies \sf I=3200\left(1.05\right)^{2} -3200

\implies \sf I=3200\left(1.1025\right) -3200

\implies \sf I=3528-3200

\implies \sf I=328

Therefore, the compound interest on the same sum for the same time at the same rate is Rs 328.

7 0
2 years ago
Kit gathered 5 eggs when she visited her cousin's farm. Altogether, the eggs weighed 225.0 grams. She wrote down the weight of e
AfilCa [17]
Easy, all you have to do is add the eggs which she got the grams for (178.7)

Then, subtract 178.7g. from 225.0g. (46.3)

Then just put the eggs according from least to greatest:
 
1st egg:43.98

2nd egg:45.02

3rd egg:45.07

4th egg:45.72

5th egg:46.3

Hope this helps!

8 0
3 years ago
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