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alexandr402 [8]
3 years ago
6

A square piece of paper has an area of x2 square units. A rectangular strip with a width of 2 units and a length of x units is c

ut off of the square piece of paper. The remaining piece of paper has an area of 120 square units.
Which equation can be used to solve for x, the side length of the original square?


x2 − 2x − 120 = 0

x2 + 2x − 120 = 0

x2 − 2x + 120 = 0

x2 + 2x + 120 = 0
Mathematics
2 answers:
vlabodo [156]3 years ago
6 0

Let

x--------> the length side of the original square paper

we know that

<u>the area of the original square paper is equal to</u>

A1=x*x=x^{2} \ units^{2}

<u>the area of the remaining piece of paper is equal to</u>

A2=x^{2} -2x\\ A2=120\ units^{2} \\ so\\ x^{2} -2x=120\\ x^{2} -2x-120=0

therefore

<u>the answer is the option </u>

x2 − 2x − 120 = 0

jekas [21]3 years ago
5 0
The answer is x² - 2x - 120 = 0
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Answer:

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Step-by-step explanation:

In this question, we have to write the slope-intercept form of the given information.

Slope intercept form is: y = mx + b

Our "m" value is our slope and our "b" value is our y-intercept.

With that knowledge, we can plug in our given information to the slope-intercept equation:

1) slope = -7/5, y-intercept = -4

Plug -7/5 to "m" and -4 to "b"

y = -7/5x - 4

2) slope = 1/5, y-intercept = -5

Plug 1/5 to "m" and -5 to "b"

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3 years ago
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Answer:

             \bold{h=\dfrac{3Q}{t-r}}

Step-by-step explanation:

\bold{r+\frac{3Q}h=t}\\-r\qquad\ -r\\\bold{\frac{3Q}h=t-r}\\\cdot h\qquad \cdot h\\ \bold{3Q=(t-r)h}\\^{\div(t-r)\quad\div(t-r)}\\\bold{\dfrac{3Q}{t-r}=h}

Or, if you mean (r+3Q)/h=t:

\bold{\frac{r+3Q}h=t}\\{}\ \ \cdot h\quad\ \cdot h\\\bold{r+3Q=ht}\\{}\ \ \div t\qquad \div t\\\bold{\dfrac{r+3Q}{t}=h}\\\\\bold{h=\dfrac{r+3Q}{t}}

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lines that are parallel have the same slope because they never intersect

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