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bearhunter [10]
3 years ago
15

A triangle has sides with lengths of 8 miles, 12 miles, and 15 miles. Is it a right triangle?

Mathematics
1 answer:
ohaa [14]3 years ago
6 0

Answer:

answer is no, not a right triangle

Step-by-step explanation:

if it is a right triangle then the sum of the square of the two smaller legs will equal the square of the longest leg

does 8² + 12² equal 15²?

64 + 144 = 225

208 ≠ 225

answer is no, not a right triangle

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-3d^2 + 10d - 8 -7d^2 + d + 6

Combine like terms to get the answer above.

Hope this helps!

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The ratio of pens to pencils in Mrs. Bosworth's desk is 4:1. What does this mean?
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Is (x+5)2 equivalent to 2x2+10x +10? Explain why or<br> why not?
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Step-by-step explanation:

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4 0
3 years ago
I need help with this Maths Homework Question.
Monica [59]

Answer:

1. 41/45.

2. x/(x^2+3x+6)

Step-by-step explanation:

1.

So first we fill the ven diagram.

There are 240 in band, so we fill that. 60 students are in both, we put that in the middle, and there are 110 people in choir.

now, since we want the probability that a student is chosen that is in band, and choir, and both. We add all this up

240 + 60 + 110 = 410.

The total possible outcome is 410, and the total outcome is 450, so the answer is

410/450 = 41/45

2.

First, to get the total outcomes, we have to add all the expressions together.

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Since that is the total outcome, we have to find the possible outcomes.

The problem wants BOTH from the 20th century and British, so it is x.

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3 0
2 years ago
Can u answer these for me with the work shown
babymother [125]

Answer:

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}

\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1

\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x

Step-by-step explanation:

Required

Simplify

Solving (1):

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}

Factorize the numerator and the denominator

\frac{x^2(x + 2) -9(x+2)}{x(x^2-x-6)}

Factor out x+2 at the numerator

\frac{(x^2 -9)(x+2)}{x(x^2-x-6)}

Express x^2 - 9 as difference of two squares

\frac{(x^2 -3^2)(x+2)}{x(x^2-x-6)}

\frac{(x -3)(x+3)(x+2)}{x(x^2-x-6)}

Expand the denominator

\frac{(x -3)(x+3)(x+2)}{x(x^2-3x+2x-6)}

Factorize

\frac{(x -3)(x+3)(x+2)}{x(x(x-3)+2(x-3))}

\frac{(x -3)(x+3)(x+2)}{x(x+2)(x-3)}

Cancel out same factors

\frac{(x+3)}{x}

Hence:

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}

Solving (2):

\frac{3x^2 - 5x - 2}{x^3 - 2x^2}

Expand the numerator and factorize the denominator

\frac{3x^2 - 6x + x - 2}{x^2(x- 2)}

Factorize the numerator

\frac{3x(x - 2) + 1(x - 2)}{x^2(x- 2)}

Factor out x - 2

\frac{(3x + 1)(x - 2)}{x^2(x- 2)}

Cancel out x - 2

\frac{3x + 1}{x^2}

Hence:

\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}

Solving (3):

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}

Express x^2 - 9 as difference of two squares

\frac{6 - 2x}{x^2 - 3^2} * \frac{15 + 5x}{4x}

Factorize all:

\frac{2(3 - x)}{(x- 3)(x+3)} * \frac{5(3 + x)}{2(2x)}

Cancel out x + 3 and 3 + x

\frac{2(3 - x)}{(x- 3)} * \frac{5}{2(2x)}

\frac{3 - x}{x- 3} * \frac{5}{2x}

Express 3 - x as -(x - 3)

\frac{-(x-3)}{x- 3} * \frac{5}{2x}\\

-1 * \frac{5}{2x}

-\frac{5}{2x}

Hence:

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}

Solving (4):

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x}

Expand x^2 - 6x + 9 and factorize 5x - 15

\frac{x^2 -3x -3x+ 9}{5(x - 3)} / \frac{5}{3-x}

Factorize

\frac{x(x -3) -3(x-3)}{5(x - 3)} / \frac{5}{3-x}

\frac{(x -3)(x-3)}{5(x - 3)} / \frac{5}{3-x}

Cancel out x - 3

\frac{(x -3)}{5} / \frac{5}{3-x}

Change / to *

\frac{(x -3)}{5} * \frac{3-x}{5}

Express 3 - x as -(x - 3)

\frac{(x -3)}{5} * \frac{-(x-3)}{5}

\frac{-(x-3)(x -3)}{5*5}

\frac{-(x-3)^2}{25}

Hence:

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}

Solving (5):

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}

Factorize the numerator and expand the denominator

\frac{x^2(x - 1) -1(x - 1)}{x^2 - x-x+1}

Factor out x - 1 at the numerator and factorize the denominator

\frac{(x^2 - 1)(x - 1)}{x(x -1)- 1(x-1)}

Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator

\frac{(x +1)(x-1)(x - 1)}{(x -1)(x-1)}

x +1

Hence:

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1

Solving (6):

\frac{9x^2 + 3x}{6x^2}

Factorize:

\frac{3x(3x + 1)}{3x(2x)}

Divide by 3x

\frac{3x + 1}{2x}

Hence:

\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}

Solving (7):

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x}

Change / to *

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}

Expand

\frac{x^2-2x-x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}

Factorize

\frac{x(x-2)-1(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}

\frac{(x-1)(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}

Cancel out x - 2 and x - 1

\frac{1}{4x} * \frac{12x^2}{x} * \frac{x}{1}

Cancel out x

\frac{1}{4x} * \frac{12x^2}{1} * \frac{1}{1}

\frac{12x^2}{4x}

3x

Hence:

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x

8 0
3 years ago
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