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2 years ago

Help! Please! Thanks

Mathematics

1 answer:

Alona [7]2 years ago

4 0

Answer:

Local maximum is at 2007

Step-by-step explanation:

Given the table which shows the number of members belonging to a community volunteer group per year. It is given that equation is cubic so let the equation be $y=ax^{3}+bx^{2}+cx+d$ , where the scale of x- axis changes in reference taken 2002 start with 0 in graph shown in attachment.

From the given data put the value of x and y and get the equation

$d=12\\a+b+c+d=45\\8a+4b+2c+d=69\\27a+9b+3c+d=85$

Solving these equations we get the value of a, b, c and d which are

$a=\frac{1}{6}, b=-5, c=\frac{227}{6}, d=12$

so, the equation is

$\frac{1}{6}x^3-5x^2+\frac{227}{6}x+12=0$

At the point of local maximum graph sudden changes it goes down, first derivative is 0. As seen in graph the point of local maximum is at he point shown in graph i.e. on the year 2007.

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The given line segment has a midpoint at (3, 1).

Katena32 [7]

Answer:

$y=\frac{1}{3}x$

Step-by-step explanation:

The given line segment has a midpoint at (3, 1) and goes through (2, 4), (3, 1), and (4, -2). We can use any two of the three points to calculate the equation of the line. Let us use the points (2, 4) and (4, -2)

Therefore the line goes through (2, 4) and (4, -2). The equation of a line passing through $(x_1,y_1)\ and\ (x_2,y_2)$ is:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Therefore the line passing through (2, 4) and (4, -2) has an equation:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y-4}{x-2}=\frac{-2-4}{4-2}\\\frac{y-4}{x-2}=\frac{-6}{2}\\y-4=x-2(-3)\\y-4=-3x+6\\y=-3x+10$

Comparing with the general equation of line: y = mx + c, the slope (m) = -3 and the intercept on the y axis (c) = 10

Two lines are said to be perpendicular if the product of their slope is -1. If the slope of line one is m1 and the slope of line 2 = m2, then the two lines are perpendicular if:

$m_1m_2=-1$ .

Therefore The slope (m2) of the perpendicular bisector of y = -3x + 10 is:

$m_1m_2=-1\\-3m_2=-1\\m_2=\frac{1}{3}$

Since it is the perpendicular bisector of the given line segment, it passes through the midpoint (3, 1). The equation of the perpendicular bisector is:

$\frac{y-y_1}{x-x_1}=m\\\frac{y-1}{x-3}=\frac{1}{3}\\ y-1= \frac{1}{3}(x-3)\\ y-1=\frac{1}{3}x-1\\y=\frac{1}{3}x$

the equation, in slope-intercept form, of the perpendicular bisector of the given line segment is $y=\frac{1}{3}x$

7 0

2 years ago

Read 2 more answers

What should you do first in order to reconcile your checkbook register?

Anarel [89]

First, you need to have your bank statement and your check register.

Then, compare all your deposits from your checkbook register to your bank statement credits to determine the amount of deposit not yet credited by the bank, which is called your Deposit in Transit.

Next, compare all your disbursements from your checkbook register to all the debits in your bank statement to determine the amount of disbursements not yet debited by the bank.

Also, you need to check for any bank charges already debited to your bank account but not yet recorded.

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2 years ago

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Katen [24]

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2 years ago

Demetrius is single but would like to take care of his mother in the event of his death. He would like to replace her income of

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1 year ago

Given the parent function f(x)=x^2 describe the graph of y=(0.2x)^2

Yanka [14]

We have the following function:
f (x) = x ^ 2
We have the following transformation:
Expansions and horizontal compressions
The graph of y = f (bx):
If 0 <b <1, the graph of y = f (x) expands horizontally by the factor of 1 / b.
Applying the transformation:
y = (0.2x) ^ 2
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b. expanded horizontally by a factor of 5

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3 years ago