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weqwewe [10]
3 years ago
7

Need help putting domain and range into set notation.The function is y = [1/2x]+1

Mathematics
1 answer:
Elanso [62]3 years ago
6 0

Answer:

Domain=2, Range=1

Step-by-step explanation:

Domain and range are like x and y, or input and output. So I would assume it would be domain=2(from the slope) and range=1(from the slope and the y-intercept) I could be very wrong tho.

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Trig, Coterminal angles... Which of the following are coterminal with -pi/4 ... (a) 7pi/4 (b)3pi/4 (c)-5pi/4) (d)-9pi/4 (e) pi/4
AveGali [126]
Correct answers are a) and d)

\alpha=- \frac{\pi}{4} 


Coterminal angles: \alpha+k\pi,\,k\in Z

- \frac{\pi}{4}+2\pi=- \frac{\pi}{4}+ \frac{8\pi}{4} = \frac{7\pi}{4}
\\
\\- \frac{\pi}{4}-2\pi=- \frac{\pi}{4}- \frac{8\pi}{4} = -\frac{9\pi}{4}
5 0
3 years ago
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2 years ago
Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

The answer is none of the choices

7 0
1 year ago
Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo
Rama09 [41]

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

u\neq 0_{R2}      

v\neq 0_{R2}

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to R^{2x2} using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]

We used the first vector ''u'' as the first column of the matrix A

We used the  second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

Det(A)=6-p

We need this determinant to be different to zero

6-p\neq 0

p\neq 6

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that p\neq 6

We can write : p ∈ IR - {6}

Notice that is p=6 ⇒

u=(3,6)

v=(1,2)

If we write 3v=3(1,2)=(3,6)=u , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.

7 0
3 years ago
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