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slega [8]
3 years ago
5

Are the following figures similar?

Mathematics
2 answers:
grin007 [14]3 years ago
7 0
C Yes; the corresponding angles are congruent
Amiraneli [1.4K]3 years ago
6 0

Answer:

B. No; the corresponding sides are not proportional.

Step-by-step explanation:

Similar figures, by definition do have all congruent corresponding angles and their legs are dilated at the same ratio. This dilation may be inversely or directly proportional. In this case we have the first condition but not the second, for we have four right angles each rectangle and its corresponding angles, but not proportional sides.

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A triangle has base 2x+1 and height 6x-3. What value of x would give an area of 240m2?
GREYUIT [131]

Answer:

The values of x which would give an area of 240m² would be:

x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}

Step-by-step explanation:

Given

The base of triangle b = 2x+1

The height of triangle h = 6x-3

The Area of the triangle A = 240 m²

The Area of the triangle has the formula

A = 1/2 × b × h

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240\:=\:\frac{1}{2}\:\left(2x+1\right)\:\times \left(6x-3\right)

480=\left(2x+1\right)\left(6x-3\right)

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Subtract 480 from both sides

12x^2-3-480=480-480

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3\left(4x^2-161\right)=0

3\left(2x+\sqrt{161}\right)\left(2x-\sqrt{161}\right)=0

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

2x+\sqrt{161}=0\quad \mathrm{or}\quad \:2x-\sqrt{161}=0

solving

2x+\sqrt{161}=0

2x=-\sqrt{161}

Divide both sides by 2

\frac{2x}{2}=\frac{-\sqrt{161}}{2}

x=-\frac{\sqrt{161}}{2}

also solving

2x-\sqrt{161}=0

2x=\sqrt{161}

Divide both sides by 2

\frac{2x}{2}=\frac{\sqrt{161}}{2}

x=\frac{\sqrt{161}}{2}

Therefore, the values of x which would give an area of 240m² would be:

x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}

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On Day 16 (in Segment 5), Ebony deposited $100 into her account. On each day after Day 16 (Segment 6) Ebony made a deposit, and
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