How to write six plus the product of w and the number two is less than or equal to -10

Mathematics

1 answer:

Marat540 [252]3 years ago

5 0

6+2w ≤ -10
This is how to write that equation.

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Explain why x and 4x - 10 are factors of the expression x(4x - 10)3 rather than terms of the expression. What are the terms of t

andriy [413]

Solution :- Let p(x) be the polynomial such that

$p(x) = x(4x-10)^3\\\text{and let f(x) be another polynomial such that }\\f(x)=4x-10\\\text{To find , Put } f(x)=0\\\Rightarrow4x-10=0\\\Rightarrow4x=10\\\Rightarrow x=5/2\\\text{Now substituting value of x in p(x),we get}\\p(x)=\frac{5}{2} .(4(5/2)-10)^3\\\Rightarrow\frac{5}{2}.(10-10)^3=0\\\text{Similarly let g(x)=x ,we get x=0 by putting g(x)=0}\\\text{now f(x)=0 for x=0}\\\text{So by Factor theorem f(x) and g(x) are the factors of polynomial p(x)}\\$

$\text{Now} p(x) = x(4x-10)^3\\=64x^4-480x^3+1200x^2-1000x\\\text{so the terms of terms of } p(x) \text{are} 64x^4,-480x^3,1200x^2,-1000x\\\text{here, by using Factor theorem x is a factor of all the terms but 4x-10 is not. }$

$\text{And the terms of the factor 4x-10 are 4x and -10}$

3 0

3 years ago

Which fraction is equivalent to 1/2 • 4/3 divided by 5/6 ?

Katen [24]

I believe the answer you are looking for is 4/5.

Solution:

1/2 • 4/3=2/3

2/3÷5/6=2/3 • 6/5

4/5

5 0

2 years ago

Solve the following equation by completing the square. 3x^2-3x-5=13

mr Goodwill [35]

we'll start off by grouping some

$\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6$

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$