PLS HELP FAST WILL GIVE BRAINLIEST!!!

The demolition durby is underway is underway

k0ka [10]

How does have anything to do with schooling or anyway helping you learn? Or how does this have anything to do with math also.

4 0

3 years ago

For the trinomial 3x^2 + 22x + 7, choose two numbers whose product is 21 and whose sum is 22

Lesechka [4]

Answer:

When a trinomial is in the form of ax2 + bx + c, where a is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form ax2 + bx + c, find two integers, r and s, whose sum is b and whose product is ac.

Step-by-step explanation:

7 0

3 years ago

Can someone please help me with this? i still have 2 more pages to do and I'm stressed out of my mind I honestly just wanna pass

melisa1 [442]

1. First we are going to find the vertex of the quadratic function $f(x)=2x^2+8x+1$ . To do it, we are going to use the vertex formula. For a quadratic function of the form $f(x)=ax^2+bx +c$ , its vertex $(h,k)$ is given by the formula $h= \frac{-b}{2a}$ ; $k=f(h)$ .

We can infer from our problem that $a=2$ and $b=8$ , sol lets replace the values in our formula:
$h= \frac{-8}{2(2)}$
$h= \frac{-8}{4}$
$h=-2$

Now, to find $k$ , we are going to evaluate the function at $h$ . In other words, we are going to replace $x$ with -2 in the function:
$k=f(-2)=2(-2)^2+8(-2)+1$
$k=f(-2)=2(4)-16+1$
$k=f(-2)=8-16+1$
$k=f(-2)=-7$
$k=-7$
So, our first point, the vertex $(h,k)$ of the parabola, is the point $(-2,-7)$ .

To find our second point, we are going to find the y-intercept of the parabola. To do it we are going to evaluate the function at zero; in other words, we are going to replace $x$ with 0:
$f(x)=2x^2+8x+1$
$f(0)=2(0)^2+(0)x+1$
$f(0)=1$
So, our second point, the y-intercept of the parabola, is the point (0,1)

We can conclude that using the vertex (-2,-7) and a second point we can graph $f(x)=2x^2+8x+1$ as shown in picture 1.

2. The vertex form of a quadratic function is given by the formula: $f(x)=a(x-h)^2+k$
where
$(h,k)$ is the vertex of the parabola.

We know from our previous point how to find the vertex of a parabola. $h= \frac{-b}{2a}$ and $k=f(h)$ , so lets find the vertex of the parabola $f(x)=x^2+6x+13$ .
$a=1$
$b=6$
$h= \frac{-6}{2(1)}$
$h=-3$
$k=f(-3)=(-3)^2+6(-3)+13$
$k=4$

Now we can use our formula to convert the quadratic function to vertex form:
$f(x)=a(x-h)^2+k$
$f(x)=1(x-(-3))^2+4$
$f(x)=(x+3)^2+4$

We can conclude that the vertex form of the quadratic function is $f(x)=(x+3)^2+4$ .

3. Remember that the x-intercepts of a quadratic function are the zeros of the function. To find the zeros of a quadratic function, we just need to set the function equal to zero (replace $f(x)$ with zero) and solve for $x$ .
$f(x)=x^2+4x-60$
$0=x^2+4x-60$
$x^2+4x-60=0$
To solve for $x$ , we need to factor our quadratic first. To do it, we are going to find two numbers that not only add up to be equal 4 but also multiply to be equal -60; those numbers are -6 and 10.
$(x-6)(x+10)=0$
Now, to find the zeros, we just need to set each factor equal to zero and solve for $x$ .
$x-6=0$ and $x+10=0$
$x=6$ and $x=-10$

We can conclude that the x-intercepts of the quadratic function $f(x)=x^2+4x-60$ are the points (0,6) and (0,-10).

4. To solve this, we are going to use function transformations and/or a graphic utility.
Function transformations.
- Translations:
We can move the graph of the function up or down by adding a constant $c$ to the y-value. If $c\ \textgreater \ 0$ , the graph moves up; if $c\ \textless \ 0$ , the graph moves down.

- We can move the graph of the function left or right by adding a constant $c$ to the x-value. If $c\ \textgreater \ 0$ , the graph moves left; if $c\ \textless \ 0$ , the graph moves right.

- Stretch and compression:
We can stretch or compress in the y-direction by multiplying the function by a constant $c$ . If $c\ \textgreater \ 1$ , we compress the graph of the function in the y-direction; if $0\ \textless \ c\ \textless \ 1$ , we stretch the graph of the function in the y-direction.

We can stretch or compress in the x-direction by multiplying $x$ by a constant $c$ . If $c\ \textgreater \ 1$ , we compress the graph of the function in the x-direction; if $0\ \textless \ c\ \textless \ 1$ , we stretch the graph of the function in the x-direction.

a. The $c$ value of $f(x)$ is 2; the $c$ value of $g(x)$ is -3. Since $c$ is added to the whole function (y-value), we have an up/down translation. To find the translation we are going to ask ourselves how much should we subtract to 2 to get -3?
$c+2=-3$
$c=-5$

Since $c\ \textless \ 0$ , we can conclude that the correct answer is: It is translated down 5 units.

b. Using a graphing utility to plot both functions (picture 2), we realize that $g(x)$ is 1 unit to the left of $f(x)$

We can conclude that the correct answer is: It is translated left 1 unit.

c. Here we have that $g(x)$ is $f(x)$ multiplied by the constant term 2. Remember that We can stretch or compress in the y-direction (vertically) by multiplying the function by a constant $c$ .

Since $c\ \textgreater \ 0$ , we can conclude that the correct answer is: It is stretched vertically by a factor of 2.

4 0

3 years ago

What is the value of g? 3g + 4 5 = 4g - 8 4?

kompoz [17]

I'm assuming this means 3g+45=4g=84.

So, just move like numbers near each other. This way, you get 4g-3g=45-84, which means g= -39.

7 0

3 years ago

Ryan has $470 in his bank account. His bank statement shows two transactions. He wrote a check for $55 and made a deposit as $30

Alchen [17]

Me das mas información?

3 0

3 years ago