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Ann [662]
3 years ago
6

Explain how you might use the properties of multiplication to solve 5 × 120 × 6.232 mentally. Then solve.

Mathematics
1 answer:
zysi [14]3 years ago
6 0

Answer:

Follows are the solution to this question:

Step-by-step explanation:

In this question, we apply the associative property that calculates the given  value, which can be defined as follows:  

Rule:

\bold{\to (A\times B)\times C=  A \times (B\times C)}

Let,

A= 5\\B=120\\C=6.232

put the value in the above rule:

Solve L.H.S part:

\to (5\times 120)\times 6.232\\\\\to (600)\times 6.232\\\\\to 3,739.2

Solve R.H.S part:

\to 5 \times (120 \times 6.232)\\\\\to 5 \times 747.84\\\\\to 3,739.2

So, L.H.S=R.H.S

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-(5m+1)<br> please help me!!!!!!!!
Art [367]

Answer:

-5m-1

Step-by-step explanation:

You have to multiply the negative sign by 5m and 1.

(-)(5m)= -5m

(-)(1)= -1

-5m - 1

4 0
3 years ago
Which equation has both 4 and -4 as possible values of y?
aleksley [76]
C) Y^2=16

Because the two numbers combined to get 16 are:

4 x 4 = 16

Or

-4 x -4 = 16
4 0
3 years ago
If 10000 is invested at an interest rate of 10 per year ,compound semiannually,find the value of the investment after the given
slega [8]

Answer:

Part a) \$17,958.56  

Part b) \$32,251.00  

Part c) \$57,918.16

Step-by-step explanation:

<u><em>The complete question is</em></u>

If $10,000 is invested at an interest rate of 10% per year, compounded semiannually, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)

a)6 years

b)12 years

c)18 years

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

Part a) 6 years

we have  

t=6\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*6}  

A=10,000(1.05)^{12}  

A=\$17,958.56  

Part b) 12 years

we have  

t=12\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*12}  

A=10,000(1.05)^{24}  

A=\$32,251.00  

Part c) 18 years

we have  

t=18\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*18}  

A=10,000(1.05)^{36}  

A=\$57,918.16

7 0
4 years ago
Enter the value of arccos(0.21) as a decimal to the nearest hundredth of a degree in the box.
8_murik_8 [283]

Answer:

77.88

Step-by-step explanation:

just did the test!

5 0
3 years ago
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Gnesinka [82]
Umm try googling it maybe I’m sorry
4 0
3 years ago
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