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3 years ago

Find the location of the absolute maximum and absolute minimum of the function on the interval [ -4,0).

1 answer:

N76 [4]3 years ago

3 0

Maximum is 0 and minimum is -4. Due to the brackets, that means that this is where the value terminates. The parentheses next to the zero means equal to

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Using the given information, give the vertex form equation for each parabola.

ollegr [7]

Answer:

Step-by-step explanation:

If you plot this point and the directrix on a coordinate plane, you can see that the directrix is 1/4 of a unit below the vertex. Since, by nature, a parabola opens in the direction opposite the directrix and "hugs" the focus, this is a positive x-squared parabola (meaning it opens upwards). The formula for this type of a parabola is, in vertex form,

$4p(y-k)=(x-h)^2$

where p is distance (in units) between the vertex and the directrix and h and k are the coordinates of the vertex. For us, p = .25, h = 7, and k = -6. Filling in our formula:

$4(.25)(y+6)=(x-7)^2$

Simplify the left side to

$1(y+6)=(x-7)^2$ which simplifies, in its entirety, to

$(x-7)^2-6=y$

3 0

3 years ago

Jovan made batches of 12 fruit bars with fruit bars in each batch. He kept 2 batches of fruit bars at home for his family and br

Galina-37 [17]

I think it’s A 200-12(b-2)

5 0

3 years ago

When the effective interest rate is 9% per annum, what is the present value of a series of 50 annual payments that start at $100

ser-zykov [4K]

Answer:

$1,109.62

Step-by-step explanation:

Let's first compute the future value FV.

In order to see the rule of formation, let's see the value (in $) for the first few years

End of year 0

1,000

End of year 1(capital + interest + new deposit)

1,000*(1.09)+10

End of year 2 (capital + interest + new deposit)

(1,000*(1.09)+10)*1.09 +10 =

$\bf 1,000*(1.09)^2+10(1+1.09)$

End of year 3 (capital + interest + new deposit)

$\bf (1,000*(1.09)^2+10(1+1.09))(1.09)+10=\\1,000*(1.09)^3+10(1+1.09+1.09^2)$

and we can see that at the end of year 50, the future value is

$\bf FV=1,000*(1.09)^{50}+10(1+1.09+(1.09)^2+...+(1.09)^{49}$

The sum

$\bf 1+1.09+(1.09)^2+...+(1.09)^{49}$

is the sum of a geometric sequence with common ratio 1.09 and is equal to

$\bf \frac{(1.09)^{50}-1}{1.09-1}=815.08356$

and the future value is then

$\bf FV=1,000*(1.09)^{50}+10*815.08356=82,508.35564$

The present value PV is

$\bf PV=\frac{FV}{(1.09)^{50}}=\frac{82508.35564}{74.35572}=1,109.616829\approx \$1,109.62$

rounded to the nearest hundredth.

5 0

3 years ago

Aiko is finding the sum (4 + 5i) + (-3 + 7i). She rewrites the sum as (-3 + 7)i + (4 + 5)i. Which statement explains the

Doss [256]

Answer:

Aiko should not have put both of the 'i' out of the brackets.

Step-by-step explanation:

As only one integer has i with it, it is not possible to take the i out of the bracket.

8 0

3 years ago

What is the interquartile range of this data? A.6 B.7 C.8 D.9

Butoxors [25]

Hello!

To find the interquartile range, subtract the value of the upper quartile from the value of the lower quartile.

33 - 25 = 8

The interquartile range is 8. Hope I helped! :3

6 0

3 years ago