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Svet_ta [14]
3 years ago
12

Determine the quadrant in which the terminal side of the given angle lies. -750°

Mathematics
1 answer:
Blababa [14]3 years ago
7 0

Answer:

Quadrant 4

Step-by-step explanation:

If the given angle was positive, then we go clockwise.

But it's negative so we go counterclockwise.

An alternative way of graphing

Quadrant 1 is 0-90°

Quadrant 2 is 90-180°

Quadrant 3 is 180-270°

Quadrant 4 is 270-360°

Subtract the given angle by 360 until no longer possible

750 - 360 = 390     390 - 360 = 30

Remember that this was originally a negative angle

Instead of going clockwise to quadrant 1, we go counterclockwise to quadrant 4, ending up at 330°

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If x < 5, then which of the following must be true?
vovangra [49]

Answer:what are the options

Step-by-step explanation:

A

B

C

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4 0
3 years ago
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7) What is the slope-intercept form of this equation? Show all your work. 6x + 2y = 12​
mariarad [96]

Answer:

3

Step-by-step explanation:

6x+2y=12

2y=6x-12

y=6x/2- 12/2

y=3x-6

7 0
3 years ago
Christopher is thinking of two positive integers. The sum of the integers is 27. When twice the first integer is added to half o
LenKa [72]

Answer:

Let x and y be the two positive integers the Christopher is thinking.

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3 years ago
The Regency Hotel has enough space at its entrance for six taxicabs to line up, wait for guests, and then load passengers. Cabs
maksim [4K]

Answer:

The appropriate solution is:

(1) 22.81 minutes

(2) 0.171

Step-by-step explanation:

According to the question, the values will be:

The service rate of guess will be:

= 5+3.5

= 8.5 \ minutes

The mean arrival rate will be:

\lambda =\frac{60}{5}

  =7.5 \ cabs/hr

The mean service rate will be:

\mu= 7.05 \ cabs/hr

(1)

The average time a cab must wait will be:

⇒ W_q=22.95-\frac{1}{7.05}

⇒       =\frac{161.798-1}{7.05}

⇒       =\frac{160.798}{7.05}

⇒       =22.81 \ minutes

(2)

The required probability will be:

⇒ P(X\geq 6)=\frac{1-2.115}{1-7.5}

⇒                  =\frac{-1.1115}{-6.5}

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6 0
3 years ago
Give three examples of equations where there would be no solution
Ronch [10]

Answer:

1) 2x+1 = 2x-4

2) 6x+2=3(2x+1)

3) -4x -5=-2(2x)+2

Step-by-step explanation:

We need to give three examples of equations where there would be no solution.

First we will define equation with no solution.

<em>If solving an equation you get the result as 3=5 but 3 is not equal to 5, so the equation has no solution.</em>

Examples of equations where there would be no solution

1) 2x+1 = 2x-4

2x-2x=-4-1

0=-1 ( false)

So, the equation has no solution.

2) 6x+2=3(2x+1)

6x+2=6x+1

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So, the equation has no solution.

3) -4x -5=-2(2x)+2

-4x-5=-4x+2

-4x+4x=2+5

0=7 (false)

So, the equation has no solution.

7 0
4 years ago
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