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3 years ago

Match the absolute value functions with their vertices

Mathematics

1 answer:

timama [110]3 years ago

7 0

Answer:

Vertex: (5, 2/3) - > f(x) = 3/5 |x - 5| + 2/3
Vertex: (-1, -3/7) - > f(x) = 1/2 |x + 1| - 3/7
Vertex: (0, -4/5) - > f(x) = 1/2 |x| - 4/5
Vertex: (2/5, 5/3) - > f(x) = 3/2 |x - 2/5| + 5/3

Step-by-step explanation:

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A plant measures 0.5 in. at the end of Week 1 and 14 in. at the end of Week 5. find the rate of change

CaHeK987 [17]

The change in growth was 14 - 0.5 = 13.5 inches.
Since it is over the course of 5 weeks, divide 13.5 by 5.
13.5/5 = 2.7
2.7 = 270%

The rate of change was 270%. Hope this helps!

8 0

3 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

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3 years ago

Timmy is selling 87 apples each apple is 1.47 each person gets 2 how many people are there

Alchen [17]

The answer is 43 people

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3 years ago

Let P be the parabola with focus (0,4) and directrix y=x.Write an equation whose graph is a parabola with a vertical directrix t

prisoha [69]

Answer:

y²=4√2.x

Step-by-step explanation:

The focus is at (0,4) and directrix is y=x or x-y =0, for a parabola P.

The distance between the focus and the directrix of the parabola P is

$\frac{ |0-4| }{\sqrt{(1)^{2}+(-1)^{2} } }$ = $\frac{4}{\sqrt{2} }$

{Since the perpendicular distance of a point (x1, y1) from the straight line ax+by+c =0 is given by $\frac{ |ax1+by1+c| }{\sqrt{a^{2}+b^{2} } }$ }

Let us assume that the equation of the parabola which is congruent with parabola P is y²=4ax

{Since the parabola has vertical directrix}

Hence, the distance between focus and the directrix is 2a = $\frac{4}{\sqrt{2} }$ , {Two parabolas are congruent when the distances between their focus and the directrix are same}

⇒ a=√2

Therefore, the equation of the parabola is y²=4√2.x (Answer)

3 0

3 years ago

7th grade math help me pleasee

zimovet [89]

Answer:

a, c, b

Step-by-step explanation:

ITS CORRECT TRUST MEEE

7 0

3 years ago

Read 2 more answers