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2 years ago

Need help please!!!!!

Mathematics

1 answer:

Yanka [14]2 years ago

6 0

Answer:

Side 1 = 8, side 2=6, side 3=3 1/3 or 10/3

Step-by-step explanation:

We can set up a proportion comparing the sides of the original triangle to the new triangle.

The original triangle has sides 12, 9, and 5. The new triangle has sides 8, e, and f (use whatever letters you like, it doesn't matter, but for me, e is the side with the middle length and f is the side with the shortest length.

We can write 3 ratios, with the length of the new triangle over the length of the old triangle.

8/12 e/9 f/5

To figure out e and f, put each ratio equal to 8/12.

8 = e Then multiply both sides of the equation by 9 and get 72 = e, 6=e.

12 9 12

8 = f Then multiply both sides of the equation by 9 and get 40 = f, 10=f.

12 5 12 3

10/3 can be rewritten as 3 1/3.

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2 years ago

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3 years ago

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Which of the following statements is false? -(-5) = 5 |-5| = -5 -|-5| = -5 -|5| = -5

TiliK225 [7]

Your question has been heard loud and clear.

Well -(-5)=5

5/-5/= -5

/-5/5= -5

So basically , the statement is false because -5(-5) not equal to -5/-5/

Thank you

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3 years ago

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Triangle ABC has vertices at A(−1 , 2), B(4, 2), and C(3, −1). Classify the triangle according to the side lengths.

Orlov [11]

Answer:

B) Isosceles

Step-by-step explanation:

The given triangle has vertices at A(-1,2), B(4,2) and C(3,-1).

We must first determine the length of the sides of the triangle, before we can classify it.

We apply the distance formula to find length of the sides.

$|AB|=\sqrt{(4--1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{(4+1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{5^2+(0)^2}$

$\Rightarrow |AB|=\sqrt{25}$

$\Rightarrow |AB|=5 units$ .

The length of side BC

$|BC|=\sqrt{(3-4)^2+(-1-2)^2}$

$\Rightarrow |BC|=\sqrt{(-1)^2+(-3)^2}$

$\Rightarrow |BC|=\sqrt{1+9}$

$\Rightarrow |BC|=\sqrt{10}$

The length of side AC

$|AC|=\sqrt{(3--1)^2+(-1-2)^2}$

We simplify to obtain;

$|AC|=\sqrt{(3+1)^2+(-3)^2}$

$\Rightarrow |AC|=\sqrt{(4)^2+(-3)^2}$

$|AC|=\sqrt{16+9}$

$|AC|=\sqrt{25}$

$|AC|=5\:units$

Since $|AC|=5\:units=|AB|$ , the given triangle is an isosceles triangle.

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3 years ago

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