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4 years ago

11

Which value is needed to create a perfect square trinomial from the expression x2 + 8x + _____?

1 answer:

Alecsey [184]4 years ago

4 0

I believe it's +16, if you complete the square you'd have (x+4)^2 - 16 + a = 0 where a is the value you're looking for, so to make this correct you need a = 16

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-1+- 4x = 11<br> How does it = this I need help pls

luda_lava [24]

X is -3 so you move -1 to the other side of the equation and change the sign to where it’s-4x=11+1 !

then u add 11+1 which is 12

so now you have -4x =12

divide both side by -4 so

-4x/4 n 12/-4

and you get -3

5 0

3 years ago

MARKING BRAINLIEST! What is 14 x 90 divided by 200

GalinKa [24]

This is through multiplication and division.

Remember your order of operations!

Answer: 6.3

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3 years ago

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Genrish500 [490]

Answer:

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3 years ago

Y=3(x-2)(x+1)(x-4) find all the zeros of the function

Akimi4 [234]

Answer:

All of the zeros would be 2, -1 and 4

Step-by-step explanation:

When you set each of the factors equal to zero solve for x. The only one you can't is 3 because $3\neq 0\\$

4 0

3 years ago

Factor this expression completely, then place the factors in the proper location on the grid.

andreev551 [17]

Notice that you have a difference of cubes:

$\dfrac{x^3}8-\dfrac{y^3}{27}=\left(\dfrac x2\right)^3-\left(\dfrac y3\right)^3$

which can be factored using the rule,

$a^3-b^3=(a-b)(a^2+ab+b^2)$

So,

$\dfrac{x^3}8-\dfrac{y^3}{27}=\left(\dfrac x2-\dfrac y3\right)\left(\dfrac{x^2}4+\dfrac{xy}6+\dfrac{y^2}9\right)$

and just to make things look nicer, we can pull out 1/6 from the first factor and 1/36 from the second to get

$\dfrac{x^3}8-\dfrac{y^3}{27}=\dfrac1{216}(3x-2y)(9x^2+6xy+4y^2)$

4 0

4 years ago