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JulijaS [17]
3 years ago
11

Need help pls and th x s​

Mathematics
1 answer:
Sedaia [141]3 years ago
4 0

Answer:

times 5

Step-by-step explanation:

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How to increase 42kg by 3%
lbvjy [14]
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.03×42=1.26

1.26kg + 42kg= 43.26

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3 years ago
Solve for p.<br><br> 2(p + 1) = 18
n200080 [17]

Answer:

p = 8

Step-by-step explanation:

Subtract 18 from behind the equal sign to cancel it out. This leaves you with 2(p+1)- 18= 0.

Next you'll need to pull out the terms to work with the beginning of the problem. 2(p+1)- 18= 0 would turn into 2p - 16  = 2(p - 8).

This would leave you with 2= 0, but that's not true so continue on to the variable.

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7 0
3 years ago
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
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