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2 years ago

100+ POINTS Go to my profile for more

Mathematics

1 answer:

VARVARA [1.3K]2 years ago

4 0

Answer:

thanks

Step-by-step explanation:

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Judy Clark went to Reel Bank. She borrowed $7,800 at a rate of 6 1/2%. The date of the loan was September 2. Judy hoped to repay

abruzzese [7]

Answer:

$7995.85

Step-by-step explanation:

We will use simple interest formula to solve our given problem.

$A=P(1+rt)$ , where,

A = Amount after t years,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

$r=6.5\%=\frac{6.5}{100}=0.065$

$t=\text{141 days}=\frac{141}{365}\text{ year}$

$A=\$78001+0.065\times \frac{141}{365})$

$A=\$7800(1+0.065\times 0.38630136986)$

$A=\$7800(1+0.025109589041)$

$A=\$7800(1.025109589041)$

$A=\$7995.85479$

$A\approx \$7995.85$

Therefore, Judy will will pay back on January 20: $7995.85.

8 0

3 years ago

The ordered pair(5,-3) is a solution to which of the following inequalities?

Dmitrij [34]

A plumber worked 26 hours a week for half a year. If his hourly wage was $40, how much did he earn during this time period?

A.$1040 B.$27,040 C.$54,080 D.$520

7 0

3 years ago

What percent of the people chose turkey or ham as their favorite sandwich? A. 6% B. 50% C. 28% D. 62%

Elenna [48]

B IS THE CORRECT ANSWER
BTWWW

7 0

2 years ago

A laser with wavelength d/8 is shining light on a double slit with slit separation 0.450 mm. This results in an interference pat

solniwko [45]

Answer:

trust no one

Step-by-step explanation:

6 0

3 years ago

SHOW YOUR WORK, PLEASE HELP!!!!!!!!!

Shkiper50 [21]

To find the inverse, we swap the variables y and x, then solve for the new y.

3a. $y=\frac{3}{x-1}$

Swapping the variables: $x=\frac{3}{y-1}$
Solving for y: $x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}$
The domain of this inverse is $x ≠ 0$ .
3b. $y=x^2-1$

Swapping: $x = y^2 - 1$
Solving for y: $y^2 = x + 1 \\ y = \sqrt{x+1}$
The domain of this inverse is $x ≥ -1$ .
3c. $y=\sqrt[3]{\frac{x-7}{3}}$
Swapping: $x=\sqrt[3]{\frac{y-7}{3}}$
Solving for y: $x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7$
The domain of this inverse is all real numbers.
4a. $y=\frac{3}{x-1}$ , $y=1+\frac{3}{x}$
$y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x$
$y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x$

4c. $y=\sqrt[3]{\frac{x-7}{3}}$ , $y=3x^3+7$
$y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x$
$y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x$

3 0

3 years ago