Plz help fast will mark the brainiest!!!

slava [35]

Both of these lines share the relationship together of being parallel.

3 0

3 years ago

Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the d

lesya [120]

Answer:

segment EH and segment E prime H prime both pass through the center of dilation.

Step-by-step explanation:

Center of dilation is point (0.1), same as H. Both, E(0,5) and H(0,1) are placed over y-axis, then E' and H' are also located at y-axis.

After dilation, H' is placed at (0,1) because it coincides with the center of dilation

Distance between E and center of dilation is 4 units, then E' should be at 4*3=12 units from the center of dilation and over y-axis. Therefore, E' is placed at (0, 13)

So, segment E'H' goes from (0,1) to (0,13), and pass through the center of dilation, like segment EH.

4 0

3 years ago

What is the cost of buying 5 kg of cheese worth<br> $9.00 per kg?

topjm [15]

$45

Hope this helps you

6 0

3 years ago

Read 2 more answers

I dont understand how to do this precalc question

Alexeev081 [22]

Answer:

x-intercept: (-0.1, 0)
Horizontal Asymptote: y = -3
Exponential <u>growth</u>

(First answer option)

Step-by-step explanation:

<u>General form of an exponential function</u>

$y=ab^x+c$

where:

a is the initial value (y-intercept).
b is the base (growth/decay factor) in decimal form:
If b > 1 then it is an increasing function.
If 0 < b < 1 then it is a decreasing function.
y=c is the horizontal asymptote.
x is the independent variable.
y is the dependent variable.

Given <u>exponential function</u>:

$y=4(10)^x-3$

<h3><u>x-intercept</u></h3>

The x-intercept is the point at which the curve crosses the x-axis, so when y = 0. To find the x-intercept, substitute y = 0 into the given equation and solve for x:

$\begin{aligned}& \textsf{Set the function to zero}:& 4(10)^x-3 &=0\\\\& \textsf{Add 3 to both sides}:& 4(10)^x &=3\\\\& \textsf{Divide both sides by 4}:& 10^x &=\dfrac{3}{4}\\\\& \textsf{Take natural logs of both sides}:& \ln 10^x &=\ln\left(\dfrac{3}{4}\right)\\\\& \textsf{Apply the power log law}:&x \ln 10 &=\ln\left(\dfrac{3}{4}\right)\\\\& \textsf{Divide both sides by }\ln 10:&x&=\dfrac{\ln\left(\dfrac{3}{4}\right)}{\ln 10} \\\\& \textsf{Simplify}:&x&=-0.1\:\:\sf(1\:d.p.)\end{aligned}$

Therefore, the x-intercept is (-0.1, 0) to the nearest tenth.

<h3><u>Asymptote</u></h3>

An <u>asymptote</u> is a line that the curve gets infinitely close to, but never touches.

The <u>parent function</u> of an <u>exponential function</u> is:

$f(x)=b^x$

As<em> </em>x approaches -∞ the function f(x) approaches zero, and as x approaches ∞ the function f(x) approaches ∞.

Therefore, there is a horizontal asymptote at y = 0.

This means that a function in the form $f(x) = ab^x+c$ always has a horizontal asymptote at y = c.

Therefore, the horizontal asymptote of the given function is y = -3.

<h3><u>Exponential Growth and Decay</u></h3>

A graph representing exponential growth will have a curve that shows an <u>increase</u> in y as x increases.

A graph representing exponential decay will have a curve that shows a <u>decrease</u> in y as x increases.

The part of an exponential function that shows the growth/decay factor is the base (b).

If b > 1 then it is an increasing function.
If 0 < b < 1 then it is a decreasing function.

The base of the given function is 10 and so this confirms that the function is increasing since 10 > 1.

Learn more about exponential functions here:

brainly.com/question/27466089

brainly.com/question/27955470

6 0

2 years ago

The general form of an parabola is 2x2−12x−3y+12=0 .

Murljashka [212]

Answer:

The standard form of the parabola is $(x-3)^2=4*\frac{3}{8}(y+2)$

Step-by-step explanation:

The standard form of a parabola is

$(x-h)^2=4p(y-k)$ .

In order to convert $2x^2-12x-3y+12=0$ into the standard form, we first separate the variables:

$2x^2-12x+3y+12=0\\\\2x^2-12x+12=3y$

we now divided both sides by 2 to remove the coefficient from $2x^2$ and get:

$x^2-6x+6=\frac{3}{2}y$ .

We complete the square on the left side by adding 3 to both sides:

$x^2-6x+6+3=\frac{3}{2}y+3$

$x^2-6x+9=\frac{3}{2}y+3$

$(x-3)^2=\frac{3}{2}y+3$

now we bring the right side into the form $4p(y-k)$ by first multiplying the equation by $\frac{2}{3}$ :

$\frac{2}{3} *(x-3)^2=\frac{2}{3} *(\frac{3}{2}y+3)\\\\\frac{2}{3} *(x-3)^2=y+2$

and then we multiplying both sides by $\frac{3}{2}$ to get

$(x-3)^2=\frac{3}{2} (y+2)$ .

Here we see that

$4p=\frac{3}{2}$

$\therefore p=\frac{3}{8}$

Thus, finally we have the equation of the parabola in the standard form:

$\boxed{(x-3)^2=4*\frac{3}{8}(y+2)}$

5 0

3 years ago