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2 years ago

( 7x + 1/2) + ( 7x - 7 and 1/2) Simplify

Mathematics

1 answer:

pishuonlain [190]2 years ago

7 0

Answer:

The answer will be 7(2x - 1)

Step-by-step explanation:

1) Convert 7 and 1/2 to improper fraction. Use this rule: a b/c = ac + b/c

$7x + \frac{1}{2} + 7x - \frac{7 \times 2 + 1}{2}$

2) Simplify 7 × 2 to 14 .

$7x + \frac{1}{2} + 7x - \frac{14 + 1}{2}$

3) Simplify 14 + 1 to 15.

$7x + \frac{1}{2} + 7x - \frac{15}{2}$

4) Collect like terms.

$(7x + 7x) + ( \frac{1}{2} - \frac{15}{2} )$

5) simplify.

$14x - 7$

6) Factor out the common term 7.

$7(2x - 1)$

Therefor, the answer is 7( 2x - 1 ).

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Which one is right..

frez [133]

Answer:

the answer would be C

Step-by-step explanation:

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2 years ago

If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha

castortr0y [4]

$6x^2+x-2=6x^2+4x-3x-2=2x(3x+2)-1(3x+2)\\\\=(3x+2)(2x-1)\\\\3x+2=0\to x=-\frac{2}{3}\\\\2x-1=0\to x=\frac{1}{2}\\\\\alpha=-\frac{2}{3};\ \beta=\frac{1}{2}\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{-\frac{2}{3}}{\frac{1}{2}}+\frac{\frac{1}{2}}{-\frac{2}{3}}=-\frac{2}{3}\cdot\frac{2}{1}-\frac{1}{2}\cdot\frac{3}{2}=-\frac{4}{3}-\frac{3}{4}\\\\=-\frac{16}{12}-\frac{9}{12}=-\frac{25}{12}=-2\frac{1}{12}$

$use\ Vieta's\ formula:\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2}{\alpha\beta}-2\\\\\alpha+\beta=\frac{-b}{a};\ \alpha\beta=\frac{c}{a}\\\\\frac{(\alpha+\beta)^2}{\alpha\beta}-2=\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-2=\frac{b^2}{a^2}\cdot\frac{a}{c}-2=\frac{b^2}{ac}-2$

$6x^2+x-2\\\\a=6;\ b=1;\ c=-2\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{1^2}{6\cdot(-2)}-2=\frac{1}{-12}-2=-2\frac{1}{12}$

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the length of the diameter of a circle inscribed in a right triangle with the length of hypotenuse c and the sum of the

jeyben [28]

Answer:

diameter = m - c

Step-by-step explanation:

In ΔABC, let ∠C be the right angle. The length of the tangents from point C to the inscribed circle are "r", the radius. Then the lengths of tangents from point A are (b-r), and those from point B have length (a-r).

The sum of the lengths of the tangents from points A and B on side "c" is ...

(b-r) +(a-r) = c

(a+b) -2r = c

Now, the problem statement defines the sum of side lengths as ...

a+b = m

and, of course, the diameter (d) is 2r, so we can rewrite the above equation as ...

m -d = c

m - c = d . . . . add d-c

The diameter of the inscribed circle is the difference between the sum of leg lengths and the hypotenuse.

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3 years ago

What is the solution of x=2+ square root x-2?

alex41 [277]

So I'm going to assume that this question is asking for non extraneous solutions, or solutions that are found in the equation and are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

$x-2=\sqrt{x-2}$

Next, square both sides:

$(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2$

Next, subtract x and add 2 to both sides of the equation:

$x^2-5x+6=0$

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

$x^2-2x-3x+6=0$

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

$x(x-2)-3(x-2)=0$

Now you can rewrite the equation as $(x-3)(x-2)=0$

Now, apply the Zero Product Property and solve for x as such:

$x-3=0\\x=3\\\\x-2=0\\x=2$

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

$2=2+\sqrt{2-2}\\2=2+\sqrt{0}\\2=2+0\\2=2\ \textsf{true}\\\\3=2+\sqrt{3-2}\\3=2+\sqrt{1}\\3=2+1\\3=3\ \textsf{true}$

Since both solutions hold true when x = 2 and x = 3, your answer is C. x = 2 or x = 3.

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3 years ago