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2 years ago

Hiii please help i’ll give brainliest if you give a correct answer ty!!

Mathematics

1 answer:

Stels [109]2 years ago

6 0

The answer is 130 because you multiply 10 x 13

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Combine the like terms

Aleksandr-060686 [28]

The answer for:

20)
6x-7+6x³-4x-10x³-4-8x - 2x²= −43−22−6−11

19)
8u+6+7u+(-10) + (-5u) + 8= 10+4

4 0

2 years ago

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Prove your work what is 1/12 of a dozen Branliest

a_sh-v [17]

Answer:

1/12 of a dozen is 1

Step-by-step explanation:

One dozen means 12. If you ask for 1/12 of 12, you multiply 12 and 1/12. You should get 1 as your answer.

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2 years ago

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Which of the following are equivalent to the expression x^1/3*x^1/2? Select all that apply

mars1129 [50]

Answer:

options 2,4 and 6 should be correct.

Step-by-step explanation:

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3 years ago

Chris opened up a new credit card and credit line with U.S bank. He is paying a 15% APR on all new purchases

Nana76 [90]

Answer:

Which one does not belong to the group

8 0

1 year ago

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

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) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

2 years ago