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katrin [286]
3 years ago
13

A gerbil cage,priced at $18, is discounted 20% for one week, and then this price is marked up 20%. is the price back to 20% expl

ain
Mathematics
1 answer:
stiks02 [169]3 years ago
6 0
No, it isn't!

first, the discount is 20 percent of 18 dollars, that is 18/5=3.6

this means that during the first week the prize is 18-3.6=14.4

now, the prize is marked up 20 % of 14.4! this is 2.88

so the prize later will be 14.4+2.88, that is 17.28, not 18

what is relevant here is what a number is a percent OF
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Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

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Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

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The values of X and Y are 30 and 11 respectively

<h3>How to determine the values of X and Y?</h3>

The figure that represents the complete question is added as an attachment

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GH = 2X -5

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From the attached parallelogram, we have:

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Substitute the known values in the above equation

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Make X the subject in X + 3 = 3Y

X = 3Y - 3

Substitute X = 3Y - 3 in 2X - 5 = 5Y

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X = 3*11 - 3

Evaluate

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