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Lorico [155]
3 years ago
8

Which pairs of triangles must be similar? Bottom question

Mathematics
1 answer:
BaLLatris [955]3 years ago
7 0

which pairs of the triangle must be similar the answer is b

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Could somebody or someone help me with this please????
Murrr4er [49]
No one knows the answer
6 0
3 years ago
colin climbs 16 feet down into tunnel and lands on the tunnel floor. then he jumps to a platform that is 2 feet above the tunnel
Ksivusya [100]

Answer:

-14

Step-by-step explanation:

-16 + 2=-14

7 0
3 years ago
Let p(x) = x^2 + 8x + 12 and q(x) = x^3 + 5x^2 - 6x. Define r(x) = p(x) \cdot q(x).
Anna [14]

The roots of the entire <em>polynomic</em> expression, that is, the product of p(x) = x^2 + 8x + 12 and q(x) = x^3 + 5x^2 - 6x, are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6.

<h3>How to solve a product of two polynomials </h3>

A value of <em>x</em> is said to be a root of the polynomial if and only if <em>r(x) =</em> 0. Let be <em>r(x) = p(x) · q(x)</em>, then we need to find the roots both for <em>p(x)</em> and <em>q(x)</em> by factoring each polynomial, the factoring is based on algebraic properties:

<em>r(x) =</em> (x + 6) · (x + 2) · x · (x² + 5 · x - 6)

<em>r(x) =</em> (x + 6) · (x + 2) · x · (x + 3) · (x + 2)

r(x) = x · (x + 2)² · (x + 3) · (x + 6)

By direct inspection, we conclude that the roots of the entire <em>polynomic</em> expression are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6. \blacksquare

To learn more on polynomials, we kindly invite to check this verified question: brainly.com/question/11536910

5 0
2 years ago
I really need help with this. I'd really appreciate anyone's help.
vekshin1

Answer:

I think it's G

Step-by-step explanation:

Since a=8 root 3=8×3=24cm

Yh so I multiplied the 24cm by 12cm

12cm×24cm=288cm squared

and the square root of 288cm is 96 root 3 cm squared

4 0
4 years ago
Y=6x 2x+3y=20 solving equation using substitution
RUDIKE [14]

Answer:

x = 1, y = 6

Step-by-step explanation:

y = 6x

2x + 3y = 20

Plug y as 6x in the second equation and solve for x.

2x + 3(6x) = 20

2x + 18x = 20

20x = 20

\frac{20x}{20}= \frac{20}{20}

x = 1

Plug x as 1 in the first equation and solve for y.

y = 6(1)

y = 6

6 0
4 years ago
Read 2 more answers
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