5 / 1/6 + 1/x+1 Math help please

Mathematics

2 answers:

dsp733 years ago

8 0

=5/ (( x+1 +6)/ 6(x+1))
= 5/ (( x+7)/ (6x + 6))
= 5(6x + 6) /x+7
= 30x + 30 / x+ 7

zubka84 [21]3 years ago

5 0

The common denomenators of the denomenaotrs is 6(x+1)
times both the fraction by (6(x+1))/(6(x+1))
we get
(5(6)(x+1))/((x+1)+6)=(30(x+1))/(x+7)=(30x+30)/(x+7)

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2x– 5(x– 3) = 2 (x– 10)

garik1379 [7]

Answer:

x-2=-3

Step-by-step explanation:

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3 0

3 years ago

Read 2 more answers

What is the quotient (125 – 8x3) ÷ (25 + 10x + 4x2)? –2x + 5 2x – 5 –2x – 5 2x + 5

juin [17]

ANSWER

$\frac{125 - 8 {x}^{3} }{ 25 + 10x + 4 {x}^{2} } = - 2x+5$

EXPLANATION

We have been given the quotient,

$\frac{125 - 8 {x}^{3} }{ 25 + 10x + 4 {x}^{2} }$
to simplify.

We need to rewrite the numerator as difference of two cubes.

$\frac{125 - 8 {x}^{3} }{ 25 + 10x + 4 {x}^{2} } = \frac{ {5}^{3} - ( {2x})^{3} }{ 25 + 10x + 4 {x}^{2} }$

We need to make use of the difference of cubes formula,

${a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )$

We now let,
$a = 5 \: and \: b = 2x$

Then the numerator becomes,

$\frac{125 - 8 {x}^{3} }{ 25 + 10x + 4 {x}^{2} } = \frac{ (5 - 2x)(25 + 5 (2x) + {(2x)}^{2}) }{ 25 + 10x + 4 {x}^{2} }$

This simplifies to,

$\frac{125 - 8 {x}^{3} }{ 25 + 10x + 4 {x}^{2} } = \frac{ (5 - 2x)(25 + 10x + 4 {x}^{2}) }{ 25 + 10x + 4 {x}^{2} }$

We cancel out common factors to obtain,

$\frac{125 - 8 {x}^{3} }{ 25 + 10x + 4 {x}^{2} } = \frac{ (5 - 2x)}{ 1}$

$\frac{125 - 8 {x}^{3} }{ 25 + 10x + 4 {x}^{2} } = 5 - 2x$