Match each justification to its corresponding step in the following solution. (Table shown above)

1213 students went on a field trip.Seven students traveled in cars and six buses were available.How many students would ride on

Alecsey [184]

Answer:201 students

Step-by-step explanation: Since there are seven kids riding in cars, you would subtract 7 from 1213 to get 1206 students. The six buses means you divide by 6 to get 201 students in each bus.

7 0

3 years ago

Which of the following describes a ratio of actual events to the number of trials in an experiment? Select all that apply. HELP

aleksklad [387]

Experiental probability is the correct choice because you will have actual data from an experiment that you are using to create the ratio of outcomes to attempts. This is the scientific and mathematical name for a tested ratio. Theoretical probability, on the other hand, is what you would expect to happen if you were to perform the test.

5 0

3 years ago

Question Completion Status:

jeka94

Answer:

$y=3x-3$

Step-by-step explanation:

The equation of the line with the slope $m$ and passing through the point $(x_0,y_0)$ is

$y-y_0=m(x-x_0)$

In your case,

$m=3\\ \\x_0=3\\ \\y_0=6$

Hence, the equation of the line is

$y-6=3(x-3)\\ \\y-6=3x-9\\ \\y=3x-9+6\\ \\y=3x-3$

3 0

3 years 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

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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

Please help it's 20 and 22 Find he value of the Variables

KIM [24]

20 x=5 I hope this helps
22.x=15

3 0

3 years ago