What is 1/10 of an income of $97.50

Match the parabolas represented by the equations with their foci.

Elenna [48]

Function 1 $f(x)=- x^{2} +4x+8$

First step: Finding when $f(x)$ is minimum/maximum
The function has a negative value $x^{2}$ hence the $f(x)$ has a maximum value which happens when $x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2$ . The foci of this parabola lies on $x=2$ .

Second step: Find the value of y-coordinate by substituting $x=2$ into $f(x)$ which give $y=- (2)^{2} +4(2)+8=12$

Third step: Find the distance of the foci from the y-coordinate
$y=- x^{2} +4x+8$ - Multiply all term by -1 to get a positive $x^{2}$
$-y= x^{2} -4x-8$ - then manipulate the constant of y to get a multiply of 4
$4(- \frac{1}{4})y= x^{2} -4x-8$
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is $12- \frac{1}{4}=11.75$

Hence the coordinate of the foci is (2, 11.75)

Function 2: $f(x)= 2x^{2}+16x+18$

The function has a positive $x^{2}$ so it has a minimum

First step - $x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4$
Second step - $y=2(-4)^{2}+16(-4)+18=-14$
Third step - Manipulating $f(x)$ to leave $x^{2}$ with constant of 1
$y=2 x^{2} +16x+18$ - Divide all terms by 2
$\frac{1}{2}y= x^{2} +8x+9$ - Manipulate the constant of y to get a multiply of 4
$4( \frac{1}{8}y= x^{2} +8x+9$

So the distance of focus from y-coordinate is $\frac{1}{8}$ to the north of $y=-14$
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: $f(x)=-2 x^{2} +5x+14$

First step: the function's maximum value happens when $x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25$
Second step: $y=-2(1.25)^{2}+5(1.25)+14=17.125$
Third step: Manipulating $f(x)$
$y=-2 x^{2} +5x+14$ - Divide all terms by -2
$-2y= x^{2} -2.5x-7$ - Manipulate coefficient of y to get a multiply of 4
$4(- \frac{1}{8})y= x^{2} -2.5x-7$
So the distance of the foci from the y-coordinate is - $\frac{1}{8}$ south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when $x=8.5$ and $y=79.25$ . The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when $x=-2.75$ and $y=-10.125$ . Manipulating coefficient of y, the distance of foci from y-coordinate is $\frac{1}{8}$ to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when $x=1.5$ and $y=9.5$ . The distance of the foci from the y-coordinate is $\frac{1}{8}$ to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

8 0

2 years ago

1/2 ÷ 2/25 what improper fraction

maw [93]

Answer:

25/4

Step-by-step explanation:

divide the 1/2 and 2/25 and thats how you get the answer:

25/4

6 0

2 years ago

Read 2 more answers

Find the missing value to the nearest hundredth. sin __ = 8/15

Sedaia [141]

Sin32.23= 8/15. This is the answer. I checked on the calculator

3 0

2 years ago

Read 2 more answers

Answer this quick I have 20 minutes<br> 25 points

e-lub [12.9K]

The answer is A. You can comment if u want an explanation :)

6 0

2 years ago

Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red

Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

Urn 1 = 2 red + 4 white = 6 chips
Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$ (4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

In the first case the probability of taking a white one is of: $\frac{2}{5}$ = 40% ( 2 whites of 5 chips)
In the second case the probability of taking a white one is of: $\frac{1}{5}$ = 20% ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66% ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$ the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06% ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$ the probability of taking a white chip from the urn two)

8 0

3 years ago