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3 years ago

Please can you fully factorise these equasions?

2 answers:

5 0

$1)\\ 36x^3-6x=6x(6x^2-1)=6x[( \sqrt{6} \cdot x)^2-1^2]=6x(x \sqrt{6} -1)(x \sqrt{6} +1)\\ \\2)\\12x^3+15x^2=3x^2(4x+3)\\ \\3)\\8x-10x^2=2x(4-5x)$

Murljashka [212]3 years ago

3 0

$36x^3-6x=6x(6x^2-1)=6x[(x\sqrt6)^2-1^2]=6x(x\sqrt6-1)(x\sqrt6+1)\\\\12x^3+15x^2=3x^2(4x+5)\\\\8x-10x^2=2x(4-5x)$

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Prove that ax+by+c=0 represent the eqn of straight line

ivann1987 [24]

If we rewrite it as y=mx+d (which can be taken from here from subtracting ax and c from both sides, then dividing b, resulting in y=(-a/b)(x)-c/b. We can then substitute -a/b for m and -c/b for d), if d=0, then we have m as a constant and as we add a specific number to y (that number being m) every time the x value increases by 1, it therefore forms a straight line. If d is not 0, then we simply add d to every single number - this is still a straight line due to that we still add a specific number to y every time x increases by 1 every single time

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3 years ago

When using the rational root theorem, which of the following is a possible root of the polynomial function below f(x)=x^3-5x^2-1

uranmaximum [27]

Answer:

$\Large \boxed{\sf \ \ 7 \ \ }$

Step-by-step explanation:

Hello, please consider the following.

The polynomial function is

$x^3-5x^2-12x+14$

The rational root theorem states that each rational solution

$x=\dfrac{p}{q}$

, written in irreducible fraction, satisfies the two following:

p is a factor of the constant term

q is a factor of the leading coefficient

In this example, the constant term is 14 and the leading coefficient is 1. It means that p is a factor of 14 and q a factor of 1.

Let's proceed with the prime factorisation of 14:

14 = 2 * 7

Finally, the possible rational roots of this expression are :

and we need to test for negative ones too

-1

-2

-7

-14

From your list, the correct answer is 7.

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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4 years ago

A tire dealer sells all-season tires for $68 each and all-terrain tires for $125 each. In one day, the sales both tires totaled

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Start by making variables to represent each tire
S=all-season tires
T=all-terrain tires
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Sub 14 into the first equation for t in order to solve for s
68s+125(14)=3926
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68s=3926-1750
68s=2176
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I hope this helps!

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3 years ago

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Draw a line through R parallel to the segment PQ

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3 years ago

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