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3 years ago

How would I find the domain and range of this graph?

Mathematics

1 answer:

matrenka [14]3 years ago

6 0

Answer:

Domain: [-3, 5]

Range: [-5, 4]

Step-by-step explanation:

The first thing is to define what is the domain and the range. The domain of a function f (x) is the set of all the values for which the function is defined, and the range of the function is the set of all the values that f takes.

In other words, the domain is the value on the "x" axis and the range is the value on the "y" axis.

In this case, both are an interval, the domain would be from -3 to 5 and in the case of the range it would be from -5 to 4.

Domain: [-3, 5]

Range: [-5, 4]

You might be interested in

Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.

fenix001 [56]

Answer:

The angle between vector $\vec{u} = 5\, \vec{i} - 8\, \vec{j}$ and $\vec{v} = 5\, \vec{i} + \, \vec{j}$ is approximately $1.21$ radians, which is equivalent to approximately $69.3^\circ$ .

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

their dot products, and
the product of their lengths.

To be precise, if $\theta$ denotes the angle between $\vec{u}$ and $\vec{v}$ (assume that $0^\circ \le \theta < 180^\circ$ or equivalently $0 \le \theta < \pi$ ,) then:

$\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ .

<h3>Dot product of the two vectors</h3>

The first component of $\vec{u}$ is $5$ and the first component of $\vec{v}$ is also

The second component of $\vec{u}$ is $(-8)$ while the second component of $\vec{v}$ is $1$ . The product of these two second components is $(-8) \times 1= (-8)$ .

The dot product of $\vec{u}$ and $\vec{v}$ will thus be:

$\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}$ .

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both $\vec{u}$ and $\vec{v}$ :

$\| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}$ .
$\| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}$ .

<h3>Angle between the two vectors</h3>

Let $\theta$ represent the angle between $\vec{u}$ and $\vec{v}$ . Apply the formula $\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ to find the cosine of this angle:

$\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}$ .

Since $\theta$ is the angle between two vectors, its value should be between $0\; \rm radians$ and $\pi \; \rm radians$ ( $0^\circ$ and $180^\circ$ .) That is: $0 \le \theta < \pi$ and $0^\circ \le \theta < 180^\circ$ . Apply the arccosine function (the inverse of the cosine function) to find the value of $\theta$ :

$\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ$ .

3 0

2 years ago

Help me please it just hard for a newbie like me

Makovka662 [10]

Hello there,
Part 1:
Perimeter of Red = 11+11+11+11
= 44
Perimeter of Blue= 6+6+6+6
= 24
Ratio: Red : Blue
44 : 24(divide by 4)
11 : 6
Part 2:
Area of Red= 11 x 11
= 121
Area of Blue= 6 x 6
= 36
Ratio: Red : Blue
121: 36

Hope this helps :))

~Top

8 0

3 years ago

Plz help no links :)

Ilya [14]

Answer:

C) f(x) = 6.25x + 3

Step-by-step explanation:

In order to know which one of the functions could produce the results in the table we simply need to substitute the number of candy bars for x in the function and solve it to see if it provides the correct total weight shown in the table. If we do this with the functions provided we can see that the only one that provides accurate results would be

f(x) = 6.25x + 3

We can input the # of candies for x and see that it provides the exact results every time as seen in the table.

f(x) = 6.25(1) + 3 = 9.25

f(x) = 6.25(2) + 3 = 15.50

f(x) = 6.25(3) + 3 = 21.75

f(x) = 6.25(4) + 3 = 28

4 0

3 years ago

Solve for the unknown values

Jet001 [13]

All you should have to do is take 180-95 to get 85 for x & then take 85+45= 130 - 180 = 50.

x=85 & y=50

5 0

2 years ago

The U.S. Weather Bureau has a station on Mauna Loa in Hawaii that has measured carbon dioxide levels since 1959. At that time, t

slavikrds [6]

The percent increase is 19.9%

3 0

3 years ago