All categories

3 years ago

Which graphs are proportional?

Mathematics

2 answers:

Komok [63]3 years ago

6 0

Answer:

Its C because it passes through the origin. I learned about this.

Mama L [17]3 years ago

4 0

Answer: C
Explanation: A proportional graph has a straight line that comes straight out of the origin (0) this is why D. cant be correct. A. Is not a straight line same with D.

You might be interested in

You earn $5 for every friendship bracelet you sell. Write and solve an equation to find the numbers of bracelets you have to sel

swat32

$85 ÷ 5 = 17

You have to sell 17 more friendship bracelets to get $85

7 0

3 years ago

Read 2 more answers

Which of the following is true of the discriminant for the graph below?

Sonbull [250]

Answer:

<u>Option C. It is zero</u>

Step-by-step explanation:

The graph represents a quadratic equation

The quadratic equation has the form ⇒a x² + b x + c

The discriminant of the quadratic equation is D = b² - 4ac

From the discriminant of the quadratic equation, we can know the type of roots of the quadratic equation.

If D > 0 ⇒ Two real roots.
If D = 0 ⇒ one real roots
If D < 0 ⇒ Two imaginary roots.

The roots of the quadratic equation are the x-intercepts of the function.

As shown at the figure, the quadratic equation has only one point of intersection with the x-axis

So, the function has only one root ⇒ D = 0

So, the discriminant of the quadratic equation = 0

<u>The answer is option C. It is zero</u>

7 0

3 years ago

LMNP is rotated 180 degrees clockwise around the origin. What are the coordinates of L?

Alona [7]

$\huge\boxed{L^\prime(0, -1)}$

To rotate a point around the origin by 180 degrees, multiply both parts of the coordinate point by $-1$ . (This means you go from $(x, y)$ to $(-x, -y)$ .)

$L(0, 1)$

$L^\prime(0*(-1), 1*(-1))\\\boxed{L^\prime(0, -1)}$

4 0

2 years ago

4+(−1 2/3) i don't get how to this cuz there aren't two fractions, just one.

kenny6666 [7]

Answer:

2 1/3

Step-by-step explanation:

4 + ( -1 2/3)

= 4 - 1 2/3

= 4 - 5/3

= 12/2 - 5/3

= 7/3

= 2 1/3

5 0

3 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago