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3 years ago

Please help don't understand

Mathematics

1 answer:

Marizza181 [45]3 years ago

3 0

I'm guess it 5 because you need to find you greatest common factors

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PLs help 50 PTS!!!!! PLEASE ILL GIVE BRAINLIEST!!!!!

Nookie1986 [14]

Answer:

$\large\boxed{y=\dfrac{1}{4}x^2-x-4}$

Step-by-step explanation:

The equation of a parabola in vertex form:

$y=a(x-h)^2+k$

(h, k) - vertex

The focus is

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of a using $k+\dfrac{1}{4a}$

k = -5

$-5+\dfrac{1}{4a}=-4$ add 5 to both sides

$\dfrac{1}{4a}=1$ multiply both sides by 4

$4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4$

$\dfrac{1}{a}=4\to a=\dfrac{1}{4}$

Substitute

$a=\dfrac{1}{4},\ h=2,\ k=-5$

to the vertex form of an equation of a parabola:

$y=\dfrac{1}{4}(x-2)^2-5$

The standard form:

$y=ax^2+bx+c$

Convert using

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

$y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5$

use the distributive property: a(b+c)=ab+ac

$y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4$

3 0

3 years ago

heeeelllllpppppp heyyy i need help asap and please try your best and solve the problem out it would be greatly appreciated becau

andreev551 [17]

Answer:

${10}^{ - 14}$

Step-by-step explanation:

$\huge\frac{ {10}^{ - 9} }{ {10}^{5} } \\ \\ \huge = {10}^{ - 9} \times {10}^{ - 5} \\ \\ \huge = {10}^{ - 9 + ( - 5)} \\ \\ \huge = {10}^{ - 9 - 5} \\ \\ \huge \orange {= {10}^{ - 14}}$

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fenix001 [56]

The answer is -7 jzzoksjsjs

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tangare [24]

Answer
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If it takes 8 inches of ribbon to make a bow,how many bows can be made from 3 feet of ribbon (1 foot = 12 inches)? Will any ribb

Aloiza [94]

First, find out how much ribbon there is.

3 x 12 = 36

Now, see how many times you can divde that by 8.

36 can only be divided by 8 four times, and there would be 4 inches of ribbon left.

So, you would be able to make 4 bows, with 4 inches of ribbon remaining.

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