All categories

3 years ago

Drag the correct graph to the box under the equation it corresponds to. HELP YOU GUYS

Mathematics

1 answer:

ra1l [238]3 years ago

5 0

Answer:

<h2>See below</h2>

Step-by-step explanation:

I can't drag and drop the graphs, but I can graph the equations shown. Then, all you will have to do is match the graphs shown the the graph that I will provide.

<h3>EQUATION 1: y = x² - 2</h3>

Graph Properties:

Opens up

Vertex is 0, -2

Axis of symmetry is x = 0

Graph photo shown in file called equation 1 graph

<h3>EQUATION 2: y = 2x²</h3>

Graph Properties:

Opens up

Vertex is 0, 0

Axis of symmetry is x = 0

Graph photo shown in file called equation 2 graph

<h3>EQUATION 3: y = (x - 2)²</h3>

Graph Properties:

Opens up

Vertex is 2, 0

Axis of symmetry is x = 2

Graph photo shown in file called equation 3 graph

I'm always happy to help :)

You might be interested in

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$

4 0

2 years ago

Marta recorded the temperature at 8 p.m. as 56°F and the temperature at 8 a.m. the next morning as 36°F. Marta assumed the tempe

Ksju [112]

1. Calculate how many hours passed between her temperature checks. 8 p.m to 8 a.m. is 12 hours.

2. Calculate the total change in temperature. 36 - 54 is -18, so the temperature went down by 18 degrees.

3. We are looking for degrees per hour, so we put derees(18) over hours (12) to get 18/12.

4. Calculate by dividing. The answer is 1.5 degrees per hour.

5 0

3 years ago

Read 2 more answers

What is the completely factored form of x4y – 4x2y – 5y?

Lady_Fox [76]

If you would like to write x^4y - 4x^2y - 5y in a completely factored form, you can do this using the following steps:

x^4y - 4x^2y - 5y = y * (x^4 - 4x^2 - 5) = y * (x^2 + 1) * (x^2 - 5)

The correct result would be <span>y * (x^2 + 1) * (x^2 - 5).</span>

4 0

3 years ago

Read 2 more answers

Please help me with a math question

tankabanditka [31]

Idk i have the same problem ugh i hate geometry

7 0

4 years ago

A cardboard carrying box has the dimensions shown below. How many square inches of cardboard is needed to make the box?

Oliga [24]

Answer:

Step-by-step explanation: