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2 years ago

To buy a computer, Raquel borrowed $3,000 at 9%

Mathematics

1 answer:

kumpel [21]2 years ago

6 0

Answer:

$4080

Step-by-step explanation:

We have the amount she will pay back, but first, we need to find the Interest accrued.

Simple Interest is given as:

$I = \frac{P * R * T }{100}$

where P = principal

R = rate

T = time taken (in years)

Therefore, the interest on $3,000 at 9% simple interest for 4 years is:

$I = \frac{3000 * 9 * 4}{100}$

I = $1080

Therefore, the amount she will pay back is:

$3000 + $1080 = $4080

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A child care center has three rooms, and the same number of children are in each room. If c children attend the center, write an

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Step-by-step explanation:

Given:

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3 years ago

The circumference of a circular hot tub at a hotel is 56.5 yards. What is the diameter of the hot tub?

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7 0

3 years ago

For the last showing of a movie, a theater sold full-price and discount tickets,the theater sold 130 full-price ticket for a tot

sveta [45]

Answer:

the cost of 1 full price ticket is $9.5

Step-by-step explanation:

For the last showing of a movie, a theater sold full-price and discount tickets,the theater sold 130 full-price ticket for a total of $1235 the total sales both types of tickets were $2395

Let x be the number of full price tickets and y be the number of discount tickets

the theater sold 130 full-price ticket for a total of $1235 the total sales

130 full price tickets sold for $1235

the cost of 1 full price ticket is

$\frac{1235}{130}=9.5$

the cost of 1 full price ticket is $9.5

4 0

3 years ago

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

7 0

3 years ago