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Arada [10]
4 years ago
5

Mark needed to buy some items for a science project. The items were priced $3.25, $0.55, $2.95, and $4.25 before tax. If everyth

ing in the store was on sale for 20% off, what was the total cost of the items he bought, before tax?
Mathematics
1 answer:
makvit [3.9K]4 years ago
3 0

The total cost was $8.80 before tax.

Step-by-step explanation:

The prices of items;

$3.25, $0.55, $2.95, and $4.25

Total price = Sum of all prices

Total price = $3.25 + $0.55 + $2.95 + $4.25

Total price = $11.00

Sale = 20% off

Discount = 20% of total price

Discount = \frac{20}{100}*11

Discount = \frac{220}{100}= \$2.20

Total cost of items bought = Total price - Discount

Total cost of items bought = 11.00 - 2.20 = $8.80

The total cost was $8.80 before tax.

Keywords: percentage, addition

Learn more about addition at:

  • brainly.com/question/11921476
  • brainly.com/question/1234767

#LearnwithBrainly

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Which number produces a rational number when added to 0.5?
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Just duplicate the number and you get a whole number which is 1

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3 years ago
What is the center of a circle whose equation is x2 + y2 + 4x – 8y + 11 = 0?
Valentin [98]

Answer:

Step-by-step explanation:

eq. of circle is x²+y²+2gx+2fy+c=0

center=(-g,-k)

given eq. is x²+y²+4x-8y+11=0

comparing

2gx=4x

g=2

2fy=-8y

f=-4

center=(-2,4)

7 0
3 years ago
Viola can work no more than 11 hours a week at her part-time job. She gets paid $7.75 per hour. Let p stand for the amount she g
Morgarella [4.7K]
The domain is 11 and the rang is $7.75.

Domain= X- values
Range= differences between the highest & lowest value
4 0
3 years ago
A 10 meter piece of wire is cut in to two pieces. One piece is 2 meters longer than the other. How long are the pieces?​
Vilka [71]

Answer:

Step-by-step explanation:

total metre = 10m

If one piece is 4m then 2m more then that = 4 + 2 = 6m

So total = 4 + 6 = 10m

7 0
3 years ago
find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6
taurus [48]

Answer:

f(x) = \frac{6}{x^2} -8 or f(x) = -\frac{6}{x^2} + 4

Step-by-step explanation:

Given

(x,y) = (1,-2) --- Point

\int\limits^6_2 {(1 + 144x^{-6})} \, dx

The arc length of a function on interval [a,b]:  \int\limits^b_a {(1 + f'(x^2))} \, dx

By comparison:

f'(x)^2 = 144x^{-6}

f'(x)^2 = \frac{144}{x^6}

Take square root of both sides

f'(x) =\± \sqrt{\frac{144}{x^6}}

f'(x) = \±\frac{12}{x^3}

Split:

f'(x) = \frac{12}{x^3} or f'(x) = -\frac{12}{x^3}

To solve fo f(x), we make use of:

f(x) = \int {f'(x) } \, dx

For: f'(x) = \frac{12}{x^3}

f(x) = \int {\frac{12}{x^3} } \, dx

Integrate:

f(x) = \frac{12}{2x^2} + c

f(x) = \frac{6}{x^2} + c

We understand that it passes through (x,y) = (1,-2).

So, we have:

-2 = \frac{6}{1^2} + c

-2 = \frac{6}{1} + c

-2 = 6 + c

Make c the subject

c = -2-6

c = -8

f(x) = \frac{6}{x^2} + c becomes

f(x) = \frac{6}{x^2} -8

For: f'(x) = -\frac{12}{x^3}

f(x) = \int {-\frac{12}{x^3} } \, dx

Integrate:

f(x) = -\frac{12}{2x^2} + c

f(x) = -\frac{6}{x^2} + c

We understand that it passes through (x,y) = (1,-2).

So, we have:

-2 = -\frac{6}{1^2} + c

-2 = -\frac{6}{1} + c

-2 = -6 + c

Make c the subject

c = -2+6

c = 4

f(x) = -\frac{6}{x^2} + c becomes

f(x) = -\frac{6}{x^2} + 4

3 0
3 years ago
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