All categories

3 years ago

Which table shows input and output values that represent a linear function?

Mathematics

1 answer:

shepuryov [24]3 years ago

4 0

Answer:

Table A

Step-by-step explanation:

In order for it to be a linear relationship, the change in the x values and y values (f(x)) needs to be consistent.

In every table, the x values change the same, so we can focus on the y values.

Notice in table A, the f(x) values go down consistently by 10 for every change in x.

All of the other tables are not consistent.

Therefore, table A is the only linear function.

You might be interested in

21. If you buy a pair of shoes for $80 and the sales tax is 3%, Find the total cost

Katarina [22]

Answer: 82.40

Step-by-step explanation: I took the quiz

6 0

2 years ago

Read 2 more answers

Help!!! Here is a screenshot of the problem.

siniylev [52]

Answer:

Step-by-step explanation:

5 0

3 years ago

An unfair coin is twice as likely to land "Heads" as "Tails". If the coin is tossed twice, what is the probability that it will

Ierofanga [76]

The probability of getting two heads on two coin tosses is 0.5 x 0.5 or 0.25. A visual representation of the toss of two coins. The Product Rule is evident from the visual representation of all possible outcomes of tossing two coins shown above.

Answer / 1/2

7 0

1 year ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

——————————

Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago

The figure below shows a shaded rectangular region inside a large rectangle

Gekata [30.6K]

Answer:

42% i hope its right if not sorry

Step-by-step explanation:

4 0

2 years ago