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2 years ago

Factorise a.x3+4.a.x

Mathematics

1 answer:

Contact [7]2 years ago

4 0

Answer:

${ax}^{3} + 4ax \\$

factorise out a and x :

${ \boxed{answer{= ax( {x}^{2} + 4) }}}\\ but \: farther \: more : \\ = ax( {x}^{2} + {2}^{2} )$

but from general factorization:

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

a » x

b » 2

therefore:

$= ax \{ {(x + 2)}^{2} - 2(x)(2) \} \\ \\ = { \boxed{ \boxed{ax( {x + 2)}^{2} - 4x }}}$

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Use the Distributive Property to rewrite the expression 5ax + 5ay as an equivalent expression.

KengaRu [80]

Answer:

5ax + 5ay = 5a(x+y)

Step-by-step explanation:

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3 years ago

Solve for y. ln(y-1)-ln2=x+ln x

alexandr402 [8]

$ln(y - 1) - ln(2) = x + ln(x)$
$e^{ln(y - 1) - ln(2)} = e^{x + ln(x)}$
$\frac{e^{ln(y - 1)}}{e^{ln(2)}} = e^{x} * e^{ln(x)}$
$\frac{y - 1}{2} = e^{x} * x$
$\frac{1}{2}y - \frac{1}{2} = xe^{x}$
$2(\frac{1}{2}y - \frac{1}{2}) = 2xe^{x}$
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3 0

3 years ago

What is one third the difference between 3/4 and 2/3?

Ivan

Answer:

One third the difference between 3/4 and 2/3 is $(\frac{1}{36} )$

Step-by-step explanation:

One third the difference of A and B = $\frac{1}{3} \times (A-B)$

Here, as given in the question:

$A = \frac{3}{4} , B = \frac{2}{3} \\\implies A - B = \frac{3}{4} - \frac{2}{3}\\=(\frac{3 \times 3 -2 \times 4}{12}) = (\frac{9-8}{12} ) = (\frac{1}{12})\\ \implies A - B = (\frac{1}{12})$

Now, one third of the difference is :

$(\frac{1}{3}) \times (\frac{1}{12}) = (\frac{1}{36})$

Hence, one third the difference between 3/4 and 2/3 is $(\frac{1}{36} )$

3 0

3 years ago

Solve for a r=x(a+b)<br><br><br> Solve for x x/2=-7

Anastaziya [24]

Answer:

a = (r/x) -b

x = -14

Step-by-step explanation:

r=x(a+b)

Divide each side by x

r/x = a+b

Subtract b

r/x -b = a

x/2=-7

Multiply each side by 2

x/2*2 = -7*2

x = -14

4 0

2 years ago

Read 2 more answers

All the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxe

AveGali [126]

Answer:

(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.

Step-by-step explanation:

Given that all the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over.

This implies that boxes are in multiples of 12

i.e. 12, or 24, or 36 or....

Let us say this as 12m for m an integer

Now additional boxes arrived =60

Total number of boxes now = $12m+60$

This when arranged in stacks of 14,nothing left out

Or this is multiple of 14

$12m+60 = 14n$ for some positive integer n.

By trial and error we find that if 60 is to be divided by 14, 24 should be added and 24 is a multiple of 12.

So we say originally 24 boxes were there now 84 boxes there.

Hence both options are right.

(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.

3 0

2 years ago

Factorise a.x3+4.a.x​

Factorise a.x3+4.a.x