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2 years ago

Please completely factor the following expression. e. 3x^2+5xy-2y^2

1 answer:

kramer2 years ago

6 0

3x^2 + 5xy - 2y^2

Simplifying:

3x^2 + 5xy + - 2y^2

Reorder Terms:
5xy +3x^2 + - 2y^2

Factor a Trinomial.
(3x + - 1y)(x + 2y)

Final Result:
(3x + - 1y)(x + 2y)

Hope this helps!!!

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Convert 131 57’ 32” to decimal degree

Arlecino [84]

Answer:

131.95888888888888

Step-by-step explanation:

Your Welcome :)

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2 years ago

3. What is wrong with the following statement? If d/e = f, then d multiplied by f = e.f should be divided by e.F and e should be

Molodets [167]

the initial statement is:

$\frac{d}{e}=f$

and then:

$d\cdot f=e$

So the second statement is false because f should be multiply by e, not by d so is option C)

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5 months ago

Please Evaluate 1/6 + 2/3

weeeeeb [17]

Answer:

Step-by-step explanation:

4/6

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2 years ago

Read 2 more answers

Which is the solution of this system of equations? y=5x−83y=x+18<br><br> y+5(3)=15

Ulleksa [173]

Answer:

y=5x−83y=x+18 (the question is wrong; it has 2 equal signs)

y+5(3)=15

y+15=15 (multiply 5*3; you get 15)

y= 15-15

y= 0

7 0

2 years ago

If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.

Lelechka [254]

Answer:

$\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }$

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

$f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)$

So we can say that

$a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}$

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

$(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab$

and we can notice that

$a^2+b^2=(a+b)^2-2ab$ so

$(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab$

it comes

$x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}$

multiply by 3

$3x^2-20x+2*16+5=3x^2-20x+37$

4 0

2 years ago