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valkas [14]
3 years ago
14

Which expression can be used to find the price of a $400 telescope after a 32% markup? Select all that apply.

Mathematics
1 answer:
emmainna [20.7K]3 years ago
7 0

Answer:

Expressions C and D correctly determine the markup price.

Step-by-step explanation:

A markup in mathematics is when an object's price is raised by a certain percentage. Therefore, because we are told that the telescope is originally $400 and is marked up by 32%, we can actually mathematically evaluate this.

When you markup a price, you multiply the original price by 1+x, where x is the percentage at which the object is marked up. However, x needs to be in decimal form for it to work out.

\frac{32}{100} = \small\boxed{0.32}

Therefore, with this new decimal, we can add 1 to it.

1 + 0.32=\small\boxed{1.32}

Finally, we multiply this result by the original price.

1.32 \times 400=\small\boxed{528}

Therefore, we can already confirm that Expression C is one way to find the markup price.

  • We can eliminate expression A because it did not use the formula 1+x. It only uses the decimal form of the percentage.
  • We can eliminate expression B because the division to convert from the percentage form to the decimal form is flawed (the expression divided by 10 instead of 100) and also forgot to add 1 to the determined value.

Now, we can test the math for expressions D and E to see if either of these will equal $528.

Expression D is basically saying to add 32% of 400 to the original price. This would work out because we marked up the original price by 32%. Therefore, we can confirm that Expression D is also an expression that can be used to determine the markup price.

400 + 400(0.32)\\400(0.32) = 128\\400 + 128 = \small\boxed{528}

Finally, we also need to check expression E. This expression is saying to add the already marked up price to 400. This would not be correct.

400 \times 400(1.32)\\400(1.32) = 528\\400 \times 528 = \small\boxed{\$ 211,200}

Therefore, it has been determined that Expression C & Expression D are both equations that will find a markup price.

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Substitute these values into equations.

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We have a system of equations now.

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2/3k - 3/4h =1

2/(4-4h) -3/4h = 1
2/(2(2-2h)) - 3/4h = 1
1/(2-2h) -3/4h - 1=0

4h/4h(2-2h) -3(2-2h)/4h(2-2h) - 4h(2-2h)/4h(2-2h) =0

(4h- 3(2-2h) - 4h(2-2h))/4h(2-2h) = 0
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Solve equation for numerator=0
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4h - 6+6h-8h+8h² =0
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4h(2-2h)≠0
1) h= 3/4  , 4*3/4(2-2*3/4)=3*(2-6)= -12 ≠0, so we can use h= 3/4
2)h=-1, 4(-1)(2-2*(-1)) =-4*4=-16 ≠0, so we can use h= -1, also.

h=3/4, then  3k = 4-4*3/4 =4 - 3=1 , 3k =1, k=1/3
h=-1, then  3k = 4-4*(-1) =8 , 3k=8, k=8/3 

So,
if h=3/4, then  k=1/3,
and if h=-1, then  k=8/3 .




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