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3 years ago

Could 14 inches, 13 inches, and 16 inches be the dimension of a triangle? Explain your answer.

Mathematics

1 answer:

marishachu [46]3 years ago

8 0

Yes, because the sum of any 2 sides of the triangle is greater than the value of the third side.

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Determine if the triangle is Right, Acute or Obtuse.

Lyrx [107]

Answer:

I think the right answer is: Acute

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3 years ago

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H = vt - 16t^2, solve for v

Sophie [7]

H=vt-16t²
vt-16t²=H
vt=16t²+H
v=(16t²+H)/t
v=16(t²/t)+H/t
v=16t+H/t

Answer: the answer would be: v=16t+H/t

4 0

3 years ago

Can’t figure this one out either

ELEN [110]

Answer: A

Step-by-step explanation:

-8 (4 - x) = 16

-32 + 8x = 16

8x = 16 + 32

8x = 48

x = 6

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3 years ago

What is the area of the triangle?<br><br> Help plz

Lunna [17]

Answer:

Step-by-step explanation:

5 0

2 years ago

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What are a) the ratio of the perimeter’s b) and the ratio of areas of the larger figure to the smaller figure?

GREYUIT [131]

a) We have been given two similar figures and we have to find ratio of perimeters of these figures.

Let a be given side of larger figure and b be the given side of smaller figure. Let $p_{1} \text{ and } p_{2}$ be the perimeters of two figures and $A_{1} \text{ and } A_{2}$ be the areas of the two figures.

We know that perimeters of two similar figures are in the ratio of corresponding sides.

$\frac{p_{1}}{p_{2}}=\frac{a}{b}$

We have been given, $a=26\text{ and }b=6$

Therefore, upon substituting these values in the above equation, we get

$\frac{p_{1}}{p_{2}}=\frac{26}{6}$

$\frac{p_{1}}{p_{2}}=\frac{13}{3}$

(b)

Further we know that areas of two similar figures are in the ratio of squares of corresponding sides.

$\frac{A_{1}}{A_{2}}=\frac{a^{2}}{b^{2}}\\
\frac{A_{1}}{A_{2}}=\frac{26^{2}}{6^{2}}\\
\frac{A_{1}}{A_{2}}=\frac{676}{36}\\
\frac{A_{1}}{A_{2}}=\frac{169}{9}\\$

Therefore, first option is the correct answer.

3 0

3 years ago

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