Would a field with a wid

Quadrilateral A'B'C'D is the image of ABCD under a dilation with a scale factor of 3 what is the length segment of BC?

irga5000 [103]

Answer:

6

Step-by-step explanation:

CD/C′D′ = 1/3

2/C D= 1/3

2 = 1/3 (CD)

6=CD

3 0

3 years ago

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write a fraction with a decimal value between 1/2 and 3/4. write both the fraction and the equivalent decimal

Irina-Kira [14]

The fraction 5/8 is a fraction between 1/2 and 3/4. Its decimal form is 0.625

6 0

4 years ago

You have a net lease for $350 per month. You are responsible for the taxes, which are $6,000 per year. What is your monthly rent

pychu [463]

Answer: 850

Step-by-step explanation: 6,000 divided by 12 = 500 + 350 = 850

7 0

3 years ago

65 yd 2 ft equals how many feet

love history [14]

It equals 197 feet

hope this helps:)

6 0

3 years ago

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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