All categories

1 year ago

Mason used a 30% coupon to buy a new computer. After the discount, the cost of the computer was $728.

Mathematics

1 answer:

shtirl [24]1 year ago

5 0

Answer:

The original price of computer is $ 1040

Solution:

Given that, Mason used a 30% coupon to buy a new computer

After the discount, the cost of the computer was $728

To find: original price of computer

From given question,

price after discount = $ 728

discount = 30 % of original price

Let "x" be the original price of computer

price after discount = original price - discount price

Step-by-step explanation:

You might be interested in

Rasheed wants to buy two new video games that cost $28.50 each. He earns money for them by mowing lawns in his neighborhood, and

myrzilka [38]

Answer:

$15.25

Step-by-step explanation:

$7.50+$4.85+$12.25 +$9.00 +$8.15 = $41.75

$28.50*2 = $57.00

So, he needs $57.00- $41.75 = $15.25 more

4 0

2 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago

Rectangle lmno is rotated 90 degrees clockwise around the origin to create l’m’n’o which statement best describes the relationsh

vova2212 [387]

Answer:

The correct option is A: they are equidistant from origin

Step-by-step explanation:

Since the rotation is performed around the fixed axis on its origin, there will not be any change to the distances between the origin and the point n or n'. The size of the rectangle has neither increased or decreased so, it cannot be possible that the distance between the origin and the point n could change after the object is rotated.

Hope that answers the question, have a great day!

4 0

2 years ago

A lake is 200 ft wide.Two boats start to move towards one another from opposite shores.In one minute the first boat goes 40 ft a

ddd [48]

Answer:

After one second, 130 ft

After two second , 60 ft

Step-by-step explanation:

Both boats are moving towards each other,

Speed of first boat = 40 ft/min.

Speed of second boat = 30 ft/min.

So, In one Second they must have covered a total distance of 40 + 30 ft

= 70 ft.

Total width of the lake = 200 ft

So, Distance remained between them after one second = 200 - 70 = 130 ft.

After two seconds , First boat will move my 80 ft and second one will move by 60 ft

So, distance between them will decrease by 140 ft.

Distance between the boats after two seconds = 60 ft.

7 0

2 years ago

Three pounds of squid can be purchased at the market for $18. Determine the equation and represent the function that defines the

olga_2 [115]

From the information given,

3 pounds = $18

Therefore, 1 pound = 18/3 = $6 (this can be treated as a constant of proportionality) such that,

Cost,C = 6*Weight, W (pounds)

Equation;

C = 6W, where C = Cost in $, and W = Weight in pounds.

7 0

2 years ago