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3 years ago

4. Describe how you can tell by looking at the graph of a function

Mathematics

1 answer:

pashok25 [27]3 years ago

3 0

Answer: The variable in the vertical axis (or y-axis) is the output

The variable in the horizontal axis (or x-axis) is the input.

Step-by-step explanation:

Usually, a graph of a function y = f(x) is represented in an X-Y coordinate axis.

An X-Y coordinate axis is conformed by two perpendicular lines, one vertical (y-axis) and one horizontal (x-axis)

The vertical axis (or the y-axis) is the output, and the horizontal axis (or the x-axis) is the input.

This method will work almost always, as is standardized that the x-axis corresponds to the input, and the y-axis corresponds to the output.

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Suppose that the breaking strength of a rope (in pounds) is normally distributed, with a mean of 100 pounds and a standard devia

EleoNora [17]

Answer:

0.961

Step-by-step explanation:

To answer this, find the area under the standard normal curve to the left of 130 pounds:

Using the function normalcdf( on a TI calculator, we get:

normalcdf(-1000, 130, 100, 17) = 0.961

8 0

3 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago

Find the area of each polygon

sweet [91]

Answer:

5 in^2

Step-by-step explanation:

Area of square: 2*2=4

Area of triangle: 2*1/2=1

4+1=5

4 0

3 years ago

A 15-foot log is split into 2 pieces. How long are the pieces if the longest piece is 1.5ft shorter than two of the shorter piec

Allisa [31]

Answer:

8.25 ft & 6.75 ft

Step-by-step explanation:

So we know it's split into two pieces, right? So normally that would be 7.5 ft each. But one is 1.5 ft longer than the other. So as a result, I divided that by 2 to add 0.75 ft to piece one and subtract 0.75 ft from piece two's length. So, the longer piece is 8.25 ft, and the shorter one is 6.75 ft.

8 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

Brilliant_brown [7]

-2x + 6y = -38

3x - 4y = 32

To solve by elimination, multiply the top equation by 3 and the bottom equation by 2.

3(-2x + 6y = -38) --> -6x + 18y = -114

2(3x - 4y = 32) --> 6x - 8y = 64

Add the equations.

-6x + 18y = -114

6x - 8y = 64

+_____________

0 + 10y = -50

10y = -50

y = -5

Substitute -5 for y into one of the original equations to find x.

3x - 4y = 32

3x - 4(-5) = 32

3x + 20 = 32

3x = 12

x = 4

Check work by plugging the x- and y-values into both of the original equations.

-2x + 6y = -38

-2(4) + 6(-5) = -38

-8 - 30 = 38

38 = 38

3x - 4y = 32

3(4) - 4(-5) = 32

12 + 20 = 32

32 = 32

Answer:

x = 4 and y = -5; (4, -5).

8 0

3 years ago

Read 2 more answers