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notsponge [240]
2 years ago
8

20 points What is the linear function equation represented by the graph?

Mathematics
1 answer:
tigry1 [53]2 years ago
4 0

Answer:

F(x) = 2/3x + 3

Step-by-step explanation:

I found this out by first starting off with the equation, f(x) = mx + b. (b is the y intercept, m is the slope.) The y intercept, where the line passes through the y axis, is 3. (f(x) = mx + 3) Now, look at rise over run, and see that the slope is 2/3, since for every one it goes over, it goes up 2/3. your final equation is f(x) = 2/3x + 3

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Answer:

214

Step-by-step explanation:

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2 years ago
A customer at Logan’s Plants pays $20.00 for a bush and $11.50 per pack of flowers. The table shows the total cost to purchase o
Volgvan

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how many expensive packs of flower is 3,150$

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1 year ago
Lin counts 5 bacteria under a microscope. She counts them again each day for four days, and finds
svetlana [45]

Answer:

Yes, but it is not linear. It is exponential :  f(x) = 5\cdot 2^{x-1}

Step-by-step explanation:

On the first day we have 5 bacteria.

On the second day we will have 5*2 = 10 bacteria.

On the third day we will have 10*2 = 5*2*2 = 5*2^2 = 20 bacteria.

On the fourth day we will have 20*2 = 5*2*2*2 = 5*2^3 = 40 bacteria.

We can see that

                                                f(x) = 5\cdot 2^{x-1},

where x is a number of days and f(x) gives us the number of bacteria.

4 0
1 year ago
Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (
musickatia [10]
(a) Suppose h_n=r^n is a solution for this recurrence, with r\neq0. Then

r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}
\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}
\implies r^4-5r^3+6r^2+4r-8=0
\implies (r-2)^3(r+1)=0\implies r=2,r=-1

So we expect a general solution of the form

h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n

With h_0=0,h_1=1,h_2=1,h_3=2, we get four equations in four unknowns:

\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}

So the particular solution to the recurrence is

h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n

(b) Let G(x)=\displaystyle\sum_{n\ge0}h_nx^n be the generating function for h_n. Multiply both sides of the recurrence by x^n and sum over all n\ge4.

\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n
\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n
G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)
G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)
(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3
G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}
G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of x^n, ideally matching the solution found in part (a).
3 0
2 years ago
Every year the state department of education gathers statistical information from all schools to award a letter grade showing th
DanielleElmas [232]
A school principal used a bar graph to send his report. He assigned the horizontal axis to the student’s name and the vertical axis to the grades. If the x-axis (the horizontal axis) is the students name and the y-axis (the vertical axis) are the grades. There has to be multiple bar-graphs per student. Otherwise the data would be incomplete.
5 0
1 year ago
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