Mathematics, 29.03.2020 03:24 ccarwile01

Simplify the following:

Answers: 1

Show answers

Answers

Answer from: zmetz405

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

Answer from: Quest

Idon’t actually know but have you asked google yet

Answer from: Quest

Ifound this on google “a skewed (non-symmetric) distribution is a distribution in which there is no such mirror-imaging. a "skewed right" distribution is one in which the tail is on the right side. a "skewed left" distribution is one in which the tail is on the left side. the above histogram is for a distribution that is skewed right.” i hope this you for what your looking for

Another question on Mathematics

Mathematics, 21.06.2019 14:30

The perimiter of a rectangle is 70 inches the legneth of the rectangle is 5 less than 3 times the width of the rectangle what is the width of the rectangle

Answers: 3

Answer

Mathematics, 21.06.2019 17:00

Asif spent $745.10 on 13 new file cabinets for his office. small file cabinets cost $43.50 and large file cabinets cost $65.95. write and solve a system of equations to find the number of small cabinets and large cabinets he purchased. determine if each statement is true or false. 1. he purchased 5 small cabinets. true or false 2. he purchased 7 large cabinets. true or false 3. he spent $527.60 on large cabinets. true or false

Answers: 2

Answer

Mathematics, 21.06.2019 19:30

Factor the polynomial 4x4 – 20x2 – 3x2 + 15 by grouping. what is the resulting expression? (4x2 + 3)(x2 – 5) (4x2 – 3)(x2 – 5) (4x2 – 5)(x2 + 3) (4x2 + 5)(x2 – 3)

Answers: 1

Answer

Mathematics, 21.06.2019 20:00

Which part of a 2-column proof is the hypothesis of the conjecture? a) given b) reasons c) prove d) statements ! i think it's b but i'm not quite

Answers: 3

Answer

You know the right answer?

Simplify the following:

...

Questions