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3 years ago

What number must be added to the expression below to complete the square? x2-5x

Mathematics

1 answer:

kati45 [8]3 years ago

3 0

Answer:

6.25

Step-by-step explanation:

(x-a)^2=x^2-2ax+a^2

2a=5

a=2.5

2.5 ^ 2 = 6.25

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3 years ago

Read 2 more answers

Make b the subject of the formula <br><img src="https://tex.z-dn.net/?f=k%20%3D%20%20%5Cfrac%7Bbrt%7D%7Bv%20-%20b%7D%20" id="Tex

balu736 [363]

Answer:

b = $\frac{kv}{k+rt}$

Step-by-step explanation:

Given

k = $\frac{brt}{v-b}$ ← multiply both sides by (v - b)

k(v - b) = brt ← distribute left side

kv - kb = brt ( subtract brt from both sides )

kv - kb - brt = 0 ( subtract kv from both sides )

- kb - brt = - kv ( multiply through by - 1 to clear the negatives )

kb + brt = kv ← factor out b from each term on the left

b(k + rt ) = kv ← divide both sides by (k + rt )

b = $\frac{kv}{k+rt}$

5 0

3 years ago

Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra

Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

$x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)$

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize $x^2-\frac13$ as yet another difference of squares:

$x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)$

And if you're working over the field of complex numbers, you can go even further. For instance,

$x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)$

But I think you'd be fine stopping at the first result,

$x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}$

6 0

3 years ago