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torisob [31]
3 years ago
7

Select all the sets of numbers that are possible values for x in the inequality, x>- 2.

Mathematics
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

{-1, 0,5}

Step-by-step explanation:

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Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:
MrMuchimi
No, it is not in the solution area for the system.

The point only works in one of the inequalities. 

It doesn't work in the inequality: y > -1/2x+ 3

We can put it in to test it:

-7 > (-1/2)3 + 3
-7 > -1.5 + 3
-7 > 1.5  This is a false statement, therefore it is not part of the solution.
5 0
4 years ago
Kendell went shopping and purchased 7 shirts, 3 pairs of pants, 2 jackets. How many different outfits are possible?
mr_godi [17]

Answer:

I dont know the answer but i have an explaination

Step-by-step explanation:

Kendall can mix the seven shirts with the two jackets which will give you fourteen possibilities for the upper part of the body, which can be mixed with 7 different options for the underpart of the body which give us a total of 98 possibilities. You need to multiply seven times two and the result that is fourteen times seven.

8 0
2 years ago
Find the value of x so that the function has the given value.<br> n(x) = -6x + 2; n(x) = 44
nalin [4]

9514 1404 393

Answer:

  x = -7

Step-by-step explanation:

Substitute the given value and solve for x

  n(x) = -6x +2

  44 = -6x + 2

  42 = -6x . . . . . subtract 2

  -7 = x . . . . . . . divide by -7

n(x) will have the value 44 for x = -7.

6 0
3 years ago
On Monday, three hundred thirteen students went on a trip to the zoo. All seven buses were filled
astra-53 [7]

Answer:

44 students were in each bus.

Step-by-step explanation:

313-5=308/7=44

8 0
4 years ago
Find all solutions in the interval from [0,2pi)<br> 2cos(3x)= -sqrt{2}
algol [13]

Solutions of 2cos(3x)= -\sqrt{2} in the interval from [0,2pi) is x =\frac{\pi}{12}  and x = \frac{23\pi}{12} .

<u>Step-by-step explanation:</u>

Find all solutions in the interval from [0,2pi)

2cos(3x)= -\sqrt{2}

⇒ 2cos(3x)= -\sqrt{2}

⇒ \frac{2cos(3x)}{2}= \frac{-\sqrt{2}}{2}

⇒ cos3x= \frac{-\sqrt{2}(\sqrt{2})}{2{\sqrt{2}}}

⇒ cos3x= \frac{-2}{2{\sqrt{2}}}

⇒ cos3x= \frac{-1}{{\sqrt{2}}}

⇒ cos^{-1}(cos3x)= cos^{-1}(\frac{-1}{{\sqrt{2}}})

⇒ 3x=\pm \frac{\pi}{4}

⇒ x=\pm \frac{\pi}{12}

Cosine General solution is :

x = \pm cos^{-1}(y)+ 2k\pi

⇒ x = \pm \frac{\pi}{12}+ 2k\pi , k is any integer .

At k=0,

⇒ x =\frac{\pi}{12} ,

At k=1,

⇒ x = - \frac{\pi}{12}+ 2\pi

⇒ x = \frac{23\pi}{12}

Therefore , Solutions of 2cos(3x)= -\sqrt{2} in the interval from [0,2pi) is x =\frac{\pi}{12}  and x = \frac{23\pi}{12} .

5 0
3 years ago
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