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3 years ago

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The math problem is listed below. Please explain how to work.

1 answer:

Nuetrik [128]3 years ago

5 0

Answer: 3/2 − i/2

Step-by-step explanation: here

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What is [(7x^3)(y^2)]^2/7 in radical form?

GuDViN [60]

Remember
$x^ \frac{m}{n} = \sqrt[n]{x^m}$
and
$(zy)^m=(z^m)(y^m)$
and
(x^m)^n=x^(mn)

so
$((7x^3)(y^2))^ \frac{2}{7} = \sqrt[7]{((7x^3)(y^2))^2}$ =
$\sqrt[7]{((7x^3)^2(y^2)^2)}$ =
$\sqrt[7]{((7^2x^6)(y^4))}$ =
$\sqrt[7]{((49x^6)(y^4))}$ =
$\sqrt[7]{49x^6y^4}$

8 0

3 years ago

The price of one share of Acme Company declined $5 per day for 4 days in a row. How much did the price of one share change in th

Ksju [112]

If the price went down $5 per day for four days straight, then it would have went down $20 total after the four days.

7 0

2 years ago

Read 2 more answers

my yearly income is 18,000. My medical insurance premium is 230 a month. Is my medical premium 9.66% less then yearly income?

JulijaS [17]

NO.....9.66% less than $18,000 is $16,261.20
230×12= $2,760

7 0

2 years 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

so

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

If the typical balance on Lucy's credit card is $650 and the interest rate (APR) on her credit card is 18%, how much in interest

mafiozo [28]

General Idea:

$Amount \; Charged \; in \; a \; Month \; = \; Monthly \; Interest \; Rate \; \times Typical \; Creditcard\; Balance\\ \\$

$Amount \; Charged \; in\; a\; month =\frac{18\%}{12} \times 650 \; = 0.015\times650=9.75$

Conclusion:

Amount of interest that Lucy to be charged in a typical month is <u>$9.75</u>

7 0

2 years ago

Read 2 more answers