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Readme [11.4K]
3 years ago
15

If the terminal point of ø is (1,0), what is tan ø?

Mathematics
1 answer:
charle [14.2K]3 years ago
4 0

Given:

The terminal point of \phi is (1,0).

To find:

The value of \tan\phi.

Solution:

If the terminal point of \theta is (x,y), then

\tan \theta=\dfrac{y}{x}

It is given that the terminal point of \phi is (1,0).

Here, x-coordinate is 1 and the y-coordinate is 0. Using the above formula, we get

\tan \phi=\dfrac{0}{1}

\tan \phi=0

Therefore, the value of \tan \phi is 0.

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0.21

Step-by-step explanation:

For independent probalities, the chance of both things happening is their product. 0.3 * 0.7 = 0.21

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Read 2 more answers
Let e1= 1 0 and e2= 0 1 ​, y1= 4 5 ​, and y2= −2 7 ​, and let​ T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps
Furkat [3]

Answer:

The image of \left[\begin{array}{c}4&-4\end{array}\right] through T is \left[\begin{array}{c}24&-8\end{array}\right]

Step-by-step explanation:

We know that T: IR^{2}  → IR^{2} is a linear transformation that maps e_{1} into y_{1} ⇒

T(e_{1})=y_{1}

And also maps e_{2} into y_{2}  ⇒

T(e_{2})=y_{2}

We need to find the image of the vector \left[\begin{array}{c}4&-4\end{array}\right]

We know that exists a matrix A from IR^{2x2} (because of how T was defined) such that :

T(x)=Ax for all x ∈ IR^{2}

We can find the matrix A by applying T to a base of the domain (IR^{2}).

Notice that we have that data :

B_{IR^{2}}= {e_{1},e_{2}}

Being B_{IR^{2}} the cannonic base of IR^{2}

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]   (Data of the problem)

T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]   (Data of the problem)

Writing the matrix A :

A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]

Now with the matrix A we can find the image of \left[\begin{array}{c}4&-4\\\end{array}\right] such as :

T(x)=Ax ⇒

T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]

We found out that the image of \left[\begin{array}{c}4&-4\end{array}\right] through T is the vector \left[\begin{array}{c}24&-8\end{array}\right]

3 0
3 years ago
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