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Sergeeva-Olga [200]
3 years ago
9

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

Mathematics
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 <em>future value FV.</em>  

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

<u>End of year 0</u>

1,000

<u>End of year 1(capital + interest + new deposit)</u>

1,000*(1.09)+10  

<u>End of year 2 (capital + interest + new deposit)</u>

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

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

<u>End of year 3 (capital + interest + new deposit)</u>

\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 <em>sum of a geometric sequence </em>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 <em>present value PV</em> 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
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