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3 years ago

!!!PLEASE HELP ASAP!!! I WILL GIVE MORR POINTS IF NEEDED AND MARK BRAINLIEST

Mathematics

1 answer:

Firlakuza [10]3 years ago

7 0

The first one shows 1 cm and 7 mm, we need to convert 7 mm to cm
1 cm + 7 mm
= 1 cm + $\frac{7}{10}$ cm
= 1 cm + 0.7 cm
= 1.7 cm
The first one is 1.7 cm

The second one shows 11 inches and 5/16 inches. Between number 11 and 12, there are 16 strips and the pointer lies on fifth strip, thus it show 5/16 inches. The second one is 11 $\frac{5}{16}$ inches.

The third one shows 13 cm and 8 mm. We need to convert mm into cm.
13 cm + 8 mm
= 13 cm + 0.8 cm
= 13.8 cm
The third one is 13.8 mm

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xz_007 [3.2K]

Answer:

1.B

2.A

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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

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4 years ago