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gulaghasi [49]
3 years ago
11

Graph h (x) = 3 cos (2x) - 1 Use 3.14 for T. Use the sine tool to graph the function. The first point must be on the midline and

the second point must be a maximum or minimum value on the graph closest to the first point.

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
6 0

Answer:

See  attachment

Step-by-step explanation:

We want to graph h(x)=3\cos(2x)-1.

We can graph this function easily by applying transformations to the parent function.

The base function is f(x)=\cos x

The transformed function has an amplitude of 3 but is shifted 1 unit down, so the minimum value is -4 and the maximum is 2.

The function is shrunk horizontally by a factor of 1/2.

The graph of the transformed function is shown in the attachment

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I’ll mark brainliest what’s 1228822882-1228822882?
xxMikexx [17]

0

Subtract

Answer is 0

Hope this helps

God bless!

4 0
2 years ago
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Determine if the following system of equations has no solutions 4x+3y= -8 -8x-6y= 16
cupoosta [38]

\begin{cases} 4x+3y=-8\\\\ -8x-6y=16 \end{cases}~\hspace{10em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill

4x+3y=-8\implies 3y=-4x-8\implies y=\cfrac{-4x-8}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3} \\\\[-0.35em] ~\dotfill\\\\ -8x-6y=16\implies -6y=8x+16\implies y=\cfrac{8x+16}{-6} \\\\\\ y=\cfrac{8}{-6}x+\cfrac{16}{-6}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3}

one simple way to tell if both equations do ever meet or have a solution is by checking their slope, notice in this case the slopes are the same for both, meaning the lines are parallel lines, however, notice both equations are really the same, namely the 2nd equation is really the 1st one in disguise.

since both equations are equal, their graph will be of one line pancaked on top of the other, and the solutions is where they meet, hell, they meet everywhere since one is on top of the other, so infinitely many solutions.

3 0
2 years ago
Oml pls helppp
USPshnik [31]

Answer:

Step-by-step explanation:

y > 2/3x - 1/5

y ≥ 3/2x + 1/5

y ≤ 2/3x + 1/5

y < 3/2x - 1/5

(Assume that the question want you to draw a graph)

Remember:

> or < is a dotted line

≥ or ≤ is a solid line

8 0
2 years ago
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a company uses two vans to transport workers from a free parking lot to the workplace between 7:00 and 9:00 am . one van has 6 m
olganol [36]
Y - x = 6
y + 2x = 57 

2y - 2x =12
y + 2x = 57 

∴3y = 69
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23 - x = 6 
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large van = 23
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4 0
3 years ago
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What point in the feasible region maximizes the objective function, 3x + y ≤ 12, x+y ≤5, x ≥0,y ≥0
forsale [732]

Step-by-step explanation:

We have to find the point in the feasible region which maximizes the objective function. To find that point first we need to graph the given inequalities to find the feasible region.

Steps to graph 3x + y ≤ 12:

First we graph 3x + y = 12 then shade the graph for ≤.

plug any value of x say x=0 and x=2 into 3x + y = 12 to find points.

plug x=0

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Now graph both points and joint them by a straight line.

test for shading.

plug any test point which is not on the graph of line like (0,0) into original inequality 3x + y ≤ 12:

3(0) + (0) ≤ 12

0 + 0 ≤ 12

0 ≤ 12

Which is true so shading will be in the direction of test point (0,0)


We can repeat same procedure to graph other inequalities.

From graph we see that ABCD is feasible region whose corner points will result into maximum or  minimu for objective function.

Since objective function is not given in the question so i will explain the process.

To find the maximum value of objective function we plug each corner point of feasible region into objective function. Whichever point gives maximum value will be the answer

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