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attashe74 [19]
2 years ago
10

A rectangular piece of paper has a width that is 3 inches less than its length. It is cut in half along a diagonal to create two

congruent right triangles with areas of 44 square inches. Which statements are true? Check all that apply.
The area of the rectangle is 88 square inches.
The equation x(x – 3) = 44 can be used to solve for the dimensions of the triangle.
The equation x2 – 3x – 88 = 0 can be used to solve for the length of the rectangle.
The triangle has a base of 11 inches and a height of 8 inches.
The rectangle has a width of 4 inches.

Mathematics
2 answers:
dybincka [34]2 years ago
5 0

Answer:

⬇⬇⬇⬇⬇⬇

⬇⬇⬇⬇⬇⬇

Step-by-step explanation:

1, 3, 4

proof below

saul85 [17]2 years ago
3 0

Answer:

I think the answer is right.isn't The rectangle width 8 and length 11 inches

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Hailey is shopping at a department store during a 20% off everything sale. She also has a coupon for $5.00 off the sale amount.
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Possible inequality:

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A circle with radius of \greenD{4\,\text{cm}}4cmstart color #1fab54, 4, start text, c, m, end text, end color #1fab54 sits insid
mario62 [17]

Answer:

The area of the shaded region is   329.87\ cm^2

Step-by-step explanation:

<u><em>The correct question is</em></u>

A circle with radius of 4cm sits inside a circle with a radius of 11cm. What is the area of the shaded region?

The shaded region is the area outside the smaller circle and inside the larger circle

we know that

To find out the shaded region subtract the area of the smaller circle from the area of the larger circle

so

A=\pi r_a^{2} -\pi r_b^{2}

simplify

A=\pi (r_a^{2} -r_b^{2})

where

r_a is the radius of the larger circle

r_b is the radius of the smaller circle

we have

r_a=11\ cm\\r_b=4\ cm

substitute

A=\pi (11^{2} -4^{2})

A=\pi (105)\ cm^2

assume

\pi=3.1416

substitute

A=(3.1416)(105)=329.87\ cm^2

7 0
2 years ago
In the derivation of the quadratic formula by completing the square, the equation (x+b over 2a)^2=-4ac+b^2 over 4a^2 is created
vredina [299]

Answer:

The result of applying the square root property of equality to this equation is x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}.

Step-by-step explanation:

Consider the provided equation.

\left(x+\dfrac{b}{2a}\right)^2=\dfrac{-4ac+b^2}{4a^2}

As the above equation is formed by perfect square trinomial so simply applying the square root property as shown:

\sqrt{(x+\dfrac{b}{2a})^2}=\pm \dfrac{\sqrt{-4ac+b^2}}{\sqrt{4a^2}}\\x+\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^2-4ac}}{2a}

Isolate the variable x.

x=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

Hence, the result of applying the square root property of equality to this equation is x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}.

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1 year ago
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yawa3891 [41]

we will find number of non-zero elements on each rows

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