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Stolb23 [73]
4 years ago
14

Stacey has a square piece of cloth. She cuts 3 inches off of the length of the square and 3 inches off of the width. The area of

the smaller square is 1/4 the area of the original square. What was the side length of the original square?
A) 1 in.
B) 2 in.
C) 6 in.
D) 12 in.
Mathematics
2 answers:
Andru [333]4 years ago
6 0
Let the side length of the original square be x, then
1/4 x^2 = (x - 3)^2
x^2 = 4(x^2 - 6x + 9) = 4x^2 - 24x + 36
3x^2 - 24x + 36 = 0
3(x^2 - 8x + 12) = 0
x^2 - 8x + 12 = 0
x^2 - 2x - 6x + 12 = 0
x(x - 2) - 6(x - 2) = 0
(x - 6)(x - 2) = 0
x - 6 = 0 or x - 2 = 0
x = 6 or x = 2
but x cannot be 2 since the side length of the smaller square will be negative.

Therefore, the side length of the original square is 6 inches.
Svetllana [295]4 years ago
3 0

Answer:

6 is the answer

Step-by-step explanation:

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Harman [31]

Answer:

1. f(x)=x^{2} and g(x)=-\sqrt{x}

2. Vertical Asymptote: x=-4

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    Because the graph will have a hole at x=-1. This is, the graph is not defined for that particular x-value.

3.  f(x)=x^{2}+4

It has two complex roots, so there is no place in the graph where it crosses the x-axis.

Step-by-step explanation:

1. If we find the composite function f[g(x)] with the functions:

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on the other hand, if we find g[f(x)] we will get the following:

g[f(x)]=-\sqrt{x^2}

which simplifies to:

g[f(x)]=-x

2.

Before finding the asymptotes, it's a good idea to factor both the numerator and denominator of the function, so we get:

f(x)=\frac{x^{2}+3x+2}{x^{2}+5x+4}

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we can now see that this function can be simplified. It is important to simplify this function so we don't confuse vertical asymptotes with holes in the graph, so when simplifying we get the following:

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

now we can set the denominator equal to zero.

x+4=0

and solve for x:

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For the horizontal asymptote, there is a rule that tells us that if the polynomials on the numerator and denominator have the same degree (the same greatest power), then the horizontal asymptote is found by dividing the lead coefficient of the polynomial on the numerator by the lead coefficient on the denominator. In this case the lead coefficients are 1 and 1 so the horizontal asymptote is:

y=\frac{1}{1}

so the horizontal asymptote is at:

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The zeros of the function are located at the x-values that will turn the numerator of the simplified fraction equal to zero, so we get:

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Take a look at the graph of the function (See attached picture)

3.

This will happen when there are no real zeros. Take for example the function:

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In this case this problem has 2 non-real zeros:

x^{2}+4=0

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x=\sqrt{-4}

so:

x=2i   or   x=-2i

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