Which of the following relations is NOT a function?

Mathematics

1 answer:

Alexxandr [17]3 years ago

5 0

Answer:

It is A.

Step-by-step explanation:

In function A we see that there are 2 ordered pairs with value 2 in the first position with y values 6 and 2. So this is not a function.

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Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

1 year ago

Your friend is selling 5 tickets to the next ice hockey game for $80. You can buy 3 tickets online for $50. Which source offers

Furkat [3]

80$ because 50 divided by 3 is 16.67 and 80 divided by 5 is 16 so yeah $80

4 0

3 years ago

Read 2 more answers

Find the median of this set of data 34 36 30 37 40 30 42 a 37 b 35 c 36 d 30

shusha [124]

Answer:

Step-by-step explanation: