PLEASE HELP. I'LL GIVE BRAINLIEST.

Mathematics

1 answer:

vova2212 [387]2 years ago

6 0

<h2>Answer:</h2>

The answer is 16.

<h2>Step-by-step explanation:</h2>

Since the 8x is positive, the equation must have y as a positive number.

Let's see this!

An easy way to complete the square if you have the constants for x2 and

But not the non-variable number is to divide the x constant, and then square it.

Example:

$\frac{8x}{2} \\\\4^2=16\\$

The last number must be 16. Let's check it.

$(x+4)^2=x^2+8x+16$

You might be interested in

Can someone please explain how to solve this.<br><br> p^-4 x -5p

solong [7]

P^-4 x -5p

In properties of exponents, if the exponent is negative then you move the variable with the negative exponent to the denominators place and it changes to positive and the numerator becomes 1.

p^-4=1/(p^4)

Also, if no exponent is present it is understood to be 1.

Keeping this in mind,
p^-4 x -5p
1/(p^4) x -5p^1 substituted p^-4 by 1/p^4 and added ^1 to -5p
-5p^1/p^4 multiplied-5p^1 to 1
-5/p^3 simplified

6 0

3 years ago

Triangle pqr has vertices p(-3 -1) q(-1,-7) and r(3,3) and points a and b are midpoints of segment pq and segment rq respectivel

Serhud [2]

Answer:

View graph

Step-by-step explanation:

Be: p(-3,-1) q(-1,-7) r(3,3)

we must find the value of the middle point a and b

$a=\frac{pq}{2}\\b=\frac{rq}{2}$

remember that distance between two points is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

pq=distance between p and q

qr =distance between q and r $pq=\sqrt{(-1+3)^{2}+(-7+1)^{2}}=\sqrt{(2)^{2}+(-6)^{2}}=\sqrt{40}\\ pq=2\sqrt{10}\\ a=\frac{pq}{2}=\frac{2\sqrt{10}}{2}=\sqrt{10}\\ coordinates\\a=\frac{pq}{2}=\frac{(-3-1,-7-1)}{2}=(-2,-4)\\\\qr=\sqrt{(3+1)^{2}+(3+7)^{2}}=\sqrt{(4)^{2}+(10)^{2}}=\sqrt{116}\\qr=2\sqrt{29}\\ b=\frac{qr}{2}=\frac{2\sqrt{29}}{2}=\sqrt{29}\\ coordinates\\b=\frac{pq}{2}=\frac{(-1+3,-7+3)}{2}=(1,-2)\\$