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kodGreya [7K]
2 years ago
15

Cos x + square root 3 = -cos x

Mathematics
1 answer:
slava [35]2 years ago
5 0

Answer:

\cos x + \sqrt{3} = -\cos x\\2\cos x = -\sqrt{3}\\\cos x =-\frac{\sqrt{3}}{2}\\x=\frac{\pi}{6}

(the other solutions can be obtained by adding 2pi as many times as you want)

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I need help i don’t understand
ch4aika [34]

9514 1404 393

Answer:

  A) J(-3, 1), ...

Step-by-step explanation:

All of the answer choices list point J' first, so it is convenient to use that as an example.

Point J on the given graph has coordinates (x, y) = (3, 2). The x-coordinate is 3 because the point is 3 units to the right of the y-axis.

The problem statement tells you to translate this point 6 units to the left. When you move it 6 units left, it will move left 3 units to the y-axis, then left 3 more units to have an x-coordinate of 3 -6 = -3. That is, each unit of movement to the left subtracts 1 from the x-coordinate. The x-coordinate of J is 3, so the final point J' will have an x-coordinate of 3 - 6 = -3.

At this point, you have enough information to make the correct answer selection. Only one answer choice has the x-coordinate of J' as -3.

__

The other coordinates are translated using similar logic. The y-coordinate of J is 2. Translating it down 1 unit subtracts 1 from the y-coordinate to make it be 2 -1 = 1. Then the coordinates of J' are (-3, 1).

We write the translation rule as ...

  (x, y) ⇒ (x -6, y -1)

This means the coordinates of each translated point have 6 subtracted from the original x-coordinate, and 1 subtracted from the original y-coordinate. The other coordinates of the figure are ...

  I(2, 4) ⇒ I'(2 -6, 4 -1) = I'(-4, 3)

  H(5, 5) ⇒ H'(5 -6, 5 -1) = H'(-1, 4)

  G(4, 1) ⇒ G'(4 -6, 1 -1) = G'(-2, 0)

3 0
3 years ago
Solve each system of linear equations albebraically
ANEK [815]

Step-by-step explanation:

1.\\\left\{\begin{array}{ccc}y=3x&(1)\\2y=6x&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\2(3x)=6x\\6x=6x\qquad\text{subtract}\ 6x\ \text{from both sides}\\6x-6x=6x-6x\\0=0\qquad\text{TRUE}\\\\Answer:\ \text{Infinitely many solutions}\\\\\left\{\begin{array}{ccc}y=3x\\x\in\mathbb{R}\end{array}\right

2.\\\left\{\begin{array}{ccc}y=2x+5&(1)\\y-2x=1&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\(2x+5)-2x=1\\2x+5-2x=1\qquad\text{combine like terms}\\(2x-2x)+5=1\\5=-1\qquad\text{FALSE}\\\\Answer:\ \text{No solution.}

3.\\\left\{\begin{array}{ccc}3x-2y=9\\-6x+4y=1&\text{divide both sides by 2}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}3x-2y=9\\-3x+2y=0.5\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad0=9.5\qquad\text{FALSE}\\\\Answer:\ \text{No solution.}

5 0
3 years ago
What is midpoint 85 90?
monitta
The midpoint of 85 and 90 is 87
8 0
3 years ago
Write an equation in standard form for the line that passes through the given points.
Agata [3.3K]

So, pretend this is your x-axis and y-axis:

                                   I

        I

   (-2,7) •  I

        I

        I       • (2, 5)

        I

        I

        I  

        I

 _________________I____________________

        I

        I

        I

TO GET FROM POINT (-2, 7) TO POINT (2, 5), WE MOVE DOWN 2 AND OVER 4, SO THE SLOPE IS -1/2. IF WE FOLLOW THAT SLOPE AND MOVE DOWN 1 AND OVER 2 FROM THE FIRST POINT OF  (-2, 7), WE WILL LAND ON A POINT LOCATED AT (0, 6), WHICH WOULD BE THE "Y-INTERCEPT". WE WERE JUST ABLE TO CALCULATE THE SLOPE OF THE LINE AND THEN USE THE SLOPE TO FIND THE INTERCEPT. SO, THE "SLOPE-INTERCEPT" FORM OF THE EQUATION FOR THIS LINE IS:

y = -1/2x + 6

TO RE-WRITE THIS IN STANDARD FORM, WE JUST WANT TO MOVE THE X VARIABLE OVER TO THE LEFT WITH THE Y VARIABLE, SO:

y = -1/2x + 6

+1/2x   + 1/2x

1/2x + y = 6 .... and that is your answer!

8 0
2 years ago
Write the equation of a line parallel to the y-axis and passing through the point<br>(-3,-4).​
Daniel [21]

Answer:

x = -3 is the answer. Writing this cause brainly requires a 20 character answer.

8 0
2 years ago
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