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Gwar [14]
3 years ago
5

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]

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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
2 years ago
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