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ser-zykov [4K]
1 year ago
7

Which is the value of this expression when p = -2 and q = -1?

Mathematics
1 answer:
Pie1 year ago
7 0

Answer:

Here is the ans...hope it helps:)

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Basically 3/8 "times" 2 

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8/2 = 4

So it will be 3/4

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is M a midpoint of line AB? AB=44, MB=22 a)maybe b)can't determine c)yes d)no (please explain how you got your answer)
Elza [17]

Answer:

Yes

Step-by-step explanation:

Yes, M is the midpoint of AB.

The length of the whole line AB is 44. So, if any midpoint is drawn on it, The two parts of the line will be equal and half of the total lines measure.

Whole line's measure = AB = 44

If any Midpoint say "M" is drawn on it, the measures of divided line would be halved.

AM = BM = 44/2 = 22

5 0
2 years ago
With the equation 2/10=4/a-3 what does a equal?
shusha [124]

Answer:

23

Step-by-step explanation:

Since 2/10 = 4/20, we know that a-3 = 20, so a = 20 + 3 = 23.

5 0
2 years ago
Will Mark Brainiest!!! Simplify the following:
xz_007 [3.2K]

Answer:

1.B

2.A

3. B

Step-by-step explanation:

1. \frac{x+5}{x^{2} + 6x +5 }

We have the denominator of the fraction as following:

x^{2} + 6x + 5 \\= x^{2} + (1 + 5)x + 5\\= x*x + 1x + 5x + 5*1\\= x ( x + 1) + 5(x + 1)\\= (x + 1) (x + 5)

As the initial one is a fraction, so that its denominator has to be different from 0.

=> (x^{2} +6x+5) ≠ 0

⇔ (x +1) (x +5) ≠ 0

⇔ (x + 1) ≠ 0; (x +5) ≠ 0

⇔ x ≠ -1; x ≠ -5

Replace it into the initial equation, we have:

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

As (x+5) ≠ 0; we divide both numerator and denominator of the fraction by (x +5)

=> \frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)} = \frac{1}{x+1}

So that \frac{x+5}{x^{2} + 6x +5 } = \frac{1}{x+1} with x ≠ 1; x ≠ -5

So that the answer is B.

2. \frac{(\frac{x^{2} -16 }{x-1} )}{x+4}

As the initial one is a fraction, so that its denominator has to be different from 0

=> x + 4 ≠ 0

=> x ≠ -4

As \frac{x^{2}-16 }{x-1} is also a fraction, so that its denominator (x-1) has to be different from 0

=> x - 1 ≠ 0

=> x ≠ 1

We have an equation: x^{2} - y^{2} = (x - y ) (x+y)

=> x^{2} - 16 = x^{2} - 4^{2} = (x -4)  (x +4)

Replace it into the initial equation, we have:

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

As (x + 4) ≠ 0 (proven above), we can divide both numerator and the denominator of the fraction by (x +4)

=> \frac{(x-4)(x+4)}{x-1} .\frac{1}{x+4} =\frac{x-4}{x-1}

So that the initial equation is equal to \frac{x-4}{x-1} with x ≠-4; x ≠1

=> So that the correct answer is A

3. \frac{x}{4x + x^{2} }

As the initial one is a fraction, so that its denominator (4x + x^2) has to be different from 0

We have:

(4x + x^2) = 4x + x.x = x ( x + 4)

So that:  (4x + x^2) ≠ 0 ⇔ x ( x + 4 ) ≠ 0

⇔ \left \{ {{x\neq 0} \atop {(x+4)\neq0 }} \right.  ⇔ \left \{ {{x\neq 0} \atop {x \neq -4 }} \right.

As (4x + x^2) = x ( x + 4) , we replace this into the initial fraction and have:

\frac{x}{4x + x^{2} } = \frac{x}{x(x+4)}

As x ≠ 0, we can divide both numerator and denominator of the fraction by x and have:

\frac{x}{x(x+4)} =\frac{x/x}{x(x+4)/x} = \frac{1}{x+4}

So that \frac{x}{4x+x^{2} }  = \frac{1}{x+4} with x ≠ 0; x ≠ -4

=> The correct answer is B

3 0
3 years ago
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