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Mathematics

1 answer:

Marta_Voda [28]2 years ago

5 0

Answer:

x ( $\frac{1}{2}$ b, $\frac{1}{2}$ c )

Step-by-step explanation:

x is the midpoint of PQ and its coordinates are the average of the coordinates of the endpoints P and Q , then

x = ( $\frac{b-0}{2}$ , $\frac{c-0}{2}$ ) = ( $\frac{1}{2}$ b, $\frac{1}{2}$ c )

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The first term of a geometric sequence is 4 and the multiplier, or ratio, is 2. what is the sum of the first 5 terms of the sequ

miss Akunina [59]

Hi,
There must have a bug on the site !!!

LoveYourselfFirst deleted your answer to the question
"Hello,Answer
=1244+8+16+32+64=124 6 minutes ago"
<span>The first term of a geometric sequence is 4 and the multiplier, or ratio, is 2. what is the sum of the first 5 terms of the sequence?
</span>
$a_1=4\\
a_2=2*a_1=8\\
a_3=2*a_2=2^2*a_1=4*4=16\\
a_4=2^3*a_1=8*4=32\\
a_5=2^4*a_1=16*4=64\\

4+8+16+32+64=124

$

<span />

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F(x) = 4√(2x³-1)<br> F'(x) =.....?

goldfiish [28.3K]

Answer:

$\large\boxed{f'(x)=\dfrac{12x^2}{\sqrt{2x^3-1}}}$

Step-by-step explanation:

$f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}$