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kkurt [141]
3 years ago
12

Simplify the expression. show your work. 22 (32 – 42)

Mathematics
2 answers:
JulijaS [17]3 years ago
8 0
If you would like to simplify the expression 22 * (32 - 42), you can do this using the following step:

22 * (32 - 42<span>) = 22 * (-10) = - 22 * 10 = - 220
</span>
The correct result would be - 220.
Alik [6]3 years ago
4 0
22(32-42)= 22(-10)
so, 22(-10)= -220
Do the math that has brackets first.
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Suppose to travel 162 miles in 3 hours .use the formula d=st to find the average speed
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2 years ago
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1) Find the lump that must be deposited today to have a future value of $ 25,000 in 5 years if funds earn 6 % componded annually
Hoochie [10]

Answer: $ 18681.45

Step-by-step explanation:

Given: Future value : FV=\$25,000

The rate of interest : r=0.06

The number of time period : t=5

The formula to calculate the future value is given by :-

\text{Future value}=P(1+i)^n, where P is the initial amount deposited.

\Rightarrow\ 25000=P(1+0.06)^5\\\\\Rightarrow\ P=\dfrac{25000}{(1.06)^5}\\\\\Rightarrow\ P=18681.4543217\approx=18681.45

Hence, the lump that must be deposit today : $ 18681.45

4 0
3 years ago
(951² - 159²)÷(7539²-357²) x (258²-852²)
Marysya12 [62]

Answer:

(879,120)÷(56709072) ×(-659340)=-10221.274

Step-by-step explanation:

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5 0
2 years ago
WILL GIVE BRAINLIEST Chang deposited $5000 into an account with a 4.8% annual interest rate, compounded monthly. Assuming that n
Mariulka [41]

Answer:

9.42 years (= 113 months)

Step-by-step explanation:

Use the compound rate interest formula:

A=P(1+\frac{r}{n})^{nt}

where:

  • A = amount
  • P = principal
  • r = interest rate (in decimal format)
  • n = number of times interest is compounded per unit t
  • t = time

Given:

  • A = $7850
  • P = $5000
  • r = 4.8% = 0.048
  • n = 12
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\implies 7850=5000(1+\frac{0.048}{12})^{12t}

\implies 7850=5000(1.004)^{12t}

\implies \dfrac{7850}{5000}=(1.004)^{12t}

\implies 1.57=(1.004)^{12t}

Take natural logs:

\implies \ln1.57=\ln(1.004)^{12t}

\implies \ln1.57=12t\ln(1.004)

\implies t=\dfrac{\ln 1.57}{12 \ln1.004}

\implies t=9.42\textsf{ years (nearest hundredth)}

\implies t=113 \textsf{ months}

5 0
1 year ago
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