Part 1: Identify key features and graph a parabola from standard form.

Four times the square of a certain number increased by 6 times the number equals 108. Find the number

vodomira [7]

-6 and 4.5
So to do this you’ll make an equation x will represent the number so 4x^2+6x=108 so we want the equation to equal 0 so we can solve it to do that you have to subtract 108 from both sides so it ends up being 4x^2+6x-108=0 we want to isolate the x onto one side and division property allows us to divide both sides by 2 the reason it’s two is because 2 is the biggest divisible number that every number in the equation is divisible by so once you divide every number by two you get 2x^2+3x-54=0 now you have to factor cause but since you only have 3x and nothing else to factor with you have to write 3x as a difference so you could do 2x^2+12x-9x-54=0 so you are complicating the problem so you can factor out 2x from the expression so 2x(x+6)- 9(x+6)=0 the 54 got factored because it’s divisible by 9 that 6 is in replace of the 54 cause if you solved it 9x6 is still 54 next factor out x+6 so (x+6) x (2x-9) =0 so one of the two have to equal 0 so right the equations separately x+6=0 minus 6 from both sides and you’ll get x to equal -6 and 2x-9=0 add 9 to both sides and you’ll get 2x=9 divide both sides by 9 and you’ll get x to equal 4.5 when you plug 4.5 and -6 for x the equation works out

3 0

3 years ago

10. Does the function represent a linear or nonlinear function? For 10 or 11

solniwko [45]

I believe their both linear because their in the correct form

4 0

2 years ago

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Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago

Simplify. +6 - 4 = please help ..........

Step2247 [10]

Answer:

2

Step-by-step explanation:

it's literally 6-4 = 2

8 0

2 years ago

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you want to create a 99 confidence interval with a margin of error of 25 assuming that the population standard deviation is equa

Ahat [919]

Answer:

The minimum sample size is 239.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Population standard deviation is equal to 1.5

This means that $\sigma = 1.5$

Margin of error of 0.25

This means that $M = 0.25$

What's the minimum size of the sample?

$M = z\frac{\sigma}{\sqrt{n}}$

$0.25 = 2.575\frac{1.5}{\sqrt{n}}$

$0.25\sqrt{n} = 2.575*1.5$

$\sqrt{n} = \frac{2.575*1.5}{0.25}$

$(\sqrt{n})^2 = (\frac{2.575*1.5}{0.25})^2$

$n = 238.7$

Rounding up:

The minimum sample size is 239.

4 0

3 years ago