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Anestetic [448]
2 years ago
5

Us jevi has three apple

Mathematics
1 answer:
Fudgin [204]2 years ago
5 0

Answer:

What do you need us to solve?

Step-by-step explanation:

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Can you guys please help me
makvit [3.9K]

Answer:

1302

Step-by-step explanation:

14 * 93 = 1302

4 0
3 years ago
Subtract (4x^2 + x + 8) - (2x^2 + 3x - 1)
Ludmilka [50]

Answer:

2x^2-2x+9

Step-by-step explanation:

5 0
3 years ago
How do you show your work for 800-374 on a number line
Sergio039 [100]

Answer:

make the numberline 800 to 0 and do bunny hop to a different part of the number line to show subtraction.

Step-by-step explanation:

7 0
3 years ago
Perform the following operations s and prove closure. Show your work.
nadezda [96]

Answer:

1. \frac{x}{x+3}+\frac{x+2}{x+5} = \frac{2x^2+10x+6}{(x+3)(x+5)}\\

2. \frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16} = \frac{1}{(x+2)(x-4)}

3. \frac{2}{x^2-9}-\frac{3x}{x^2-5x+6} = \frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}

4. \frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3} = \frac{1}{(x-2)(x-4)}

Step-by-step explanation:

1. \frac{x}{x+3}+\frac{x+2}{x+5}

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. \frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

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

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

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

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

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

=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}

Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.

4. \frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}

Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

\frac{x+4}{(x-2)(x+3)}\div\frac{(x-4)(x+4)}{x+3}

Converting ÷ sign into multiplication we will take reciprocal of the second term

=\frac{x+4}{(x-2)(x+3)}*\frac{x+3}{(x-4)(x+4)}\\=\frac{1}{(x-2)(x-4)}

Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.

5 0
3 years ago
Explain? How to slove it
Neko [114]
The answer is F because you use the square's measurements as the base of each triangle then multiply each triangle then divide by 2. The multiply by 4 because there are four triangles and multiply 12 by 12 to get 144 because it is the base and the height of the square.
6 0
3 years ago
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