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2 years ago

13. A pair of shoes sell for $27 per pair. There is a sale tomorrow on shoes offering two pairs for $45.

Mathematics

1 answer:

agasfer [191]2 years ago

8 0

Answer:

$72

Step-by-step explanation:

The sale offers 2 pair of shoes for $45, but the price is same for 1 pair of shoes.

The cost of 3 pair of shoes

= The sale price of 2 pair of shoes + Regular price of 1 pair of shoes

= $45 + $27

= $72

So, 3 pairs of shoes will cost $72 today.

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2 years ago

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What is the common ratio of the following geometric sequence 5/12, 1/4,3/20,9/100,27/500

suter [353]

Answer:

3/5

Step-by-step explanation

Ratio =second term divided inti first term a2/a1

a1=5/12 a2=1/4

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2 years ago

Describe the position of the function below relative to the graph of the indicated basic function.

matrenka [14]

From the given the option moved left 1 unit; reflected across the x axis is the correct.

what is function?

A function from a set X to a set Y in mathematics assigns each element of X exactly one element of Y. The set X is known as the function's domain, and the set Y is known as the function's codomain.

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After shifting one unit right it becomes, $2^x^+^1$ .

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Hence, from the given the option moved left 1 unit; reflected across the x axis is the correct.

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10 months ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

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Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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3 years ago