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maw [93]
3 years ago
5

4-15 all this is overdue and I cant understand anything

Mathematics
1 answer:
attashe74 [19]3 years ago
5 0

I'm not 100% sure how to do this, but it would help if you had any notes/online notes/textbook that shows how to do it (also to see if what I did is the same as the notes)


4. B<u>reak-even point</u> (It's basically the intersection point. Or the point when the two lines meet/the point that is the same for both equations)

I plugged in numbers for x into both equations, until C and R were the same number.  

I had x = 1, so I plugged in 1 for x into both equations:

C = 15x + 150                      C = 15(1) + 150 = 15 + 150 = 165

R = 45x                                R = 45(1) = 45

Then I had x = 2, then x = 3, and so on until C and R were the same number (meaning they had the same point), which was (5, 225)


Idk how you are suppose to check the solution because there are multiple ways to do this. I doubt this, but one option is plugging (5, 225) back into the equations:

C = 15x + 150                                                                R = 45x

225 = 15(5) + 150                                                     225 = 45(5)

225 = 75 + 150                                                         225 = 225

225 = 225


5. (I made another sort of table? because I didn't know what kind you were suppose to use)   Do the same as #4, and plug in numbers for x until C and R were the same number


6. Do the same as #4   (If you have any questions or information I should know then let me know)


7.   y = mx + b  

"m" is the slope, "b" is the y-intercept [the y value when x = 0   (0, y)]

The y-intercepts are -2 and 13, so the graph is B. Now you need to estimate the solution(point of intersection) by looking at the graph. It looks around (6,7).

Now you need to check the solution by plugging the point into both equations.

y = 1.5x - 2

7 = 1.5(6) - 2

7 = 9 - 2

7 = 7


y = -x + 13

7 = -6 + 13

7 = 7                 The point is the solution


8. The graph is A. Do the same as #7 and estimate a point. I will say around (2.5,6.5)  And check this by plugging the point into both equations

y = x + 4

6.5 = 2.5 + 4

6.5 = 6.5


y = 3x - 1

6.5 = 3(2.5) - 1

6.5 = 7.5 - 1

6.5 = 6.5        The point is the solution


9. Graph C, do the same as #7 and #8


10. y = mx + b

"b" is (0,b) or (0,y)

"m" is the slope

(slope=\frac{rise}{run}  

Rise is the number of units you go up(+) or down(-)

Run is the number of units you go to the right)


You need to graph the lines of each equation, and find the point of intersection.


11-15. Do the same as #10

If "y" is not by itself, get it by itself.

#13:

x + y = 27    Subtract x on both sides to get "y" by itself

y = 27 - x

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Find , , and if and terminates in quadrant .
ioda

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

STEP - BY - STEP EXPLANATION

What to find?

• sin2x

,

• cos2x

,

• tan2x

Given:

tanx = 2/3 = opposite / adjacent

We need to first make a sketch of the given problem.

Let h be the hypotenuse.

We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.

Using the Pythagoras theorem;

hypotenuse² = opposite² + adjacent²

h² = 2² + 3²

h² = 4 + 9

h² =13

Take the square root of both-side of the equation.

h =√13

This implies that hypotenuse = √13

We can now proceed to find the values of ainx and cosx.

Using the trigonometric ratio;

\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}}\cos x=\frac{adjacent}{\text{hypotenuse}}=\frac{3}{\sqrt[]{13}}

And we know that tanx =2/3

From the trigonometric identity;

sin 2x = 2sinxcosx

Substitute the value of sinx , cosx and then simplify.

\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})=\frac{12}{13}

Hence, sin2x = 12/13

cos2x = cos²x - sin²x

Substitute the value of cosx, sinx and simplify.

\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\  \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}

Hence, cos2x = 5/13

tan2x = 2tanx / 1- tan²x

\tan 2x=\frac{2\tan x}{1-\tan ^2x}=\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^2}=\frac{\frac{4}{3}}{1-\frac{4}{9}}=\frac{\frac{4}{3}}{\frac{9-4}{9}}=\frac{\frac{4}{3}}{\frac{5}{9}}=\frac{4}{3}\times\frac{9}{5}=\frac{4}{1}\times\frac{3}{5}=\frac{12}{5}

OR

\tan 2x=\frac{\sin 2x}{\cos 2x}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}

Hence, tan2x = 12/5

Therefore,

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

3 0
1 year ago
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