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Gelneren [198K]
4 years ago
12

What is the solution set for the inequality?

Mathematics
2 answers:
irakobra [83]4 years ago
8 0

Answer:

tutututkknñnnñnnncxmdksufjskskckd sorry

Reil [10]4 years ago
7 0
Like 6 or 7 might be 8 or maybe even 9
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What is the volume of this sphere? Need help asap. Will Mark Brainliest if answered correctly . ​
Romashka-Z-Leto [24]

Answer:

288

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3 years ago
What is the distance between (0,3) and (4,6)
Genrish500 [490]

Answer: 5 units

Step-by-step explanation:

7 0
3 years ago
Help, ill mark brainliest
ASHA 777 [7]

Answer: FIRST OPTION.

Step-by-step explanation:

First, it is important to remember that the Slope-Intercept form of the equation of a line is the shown below:

y=mx+b

Where "m" is the slope of the line and "b" is the y-intercept.

By definiton, given a System of Linear equations, if they are exactly the same line, then the System of equations have Infinely many solutions.

In this case you have the following System of Linear equations given in the exercise:

\left \{ {{y=-2x+5} \atop {y=ax+b}} \right.

So, since you need the system has Infinite solutions, you know that the slope and the y-intercept of both lines must be equal.

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a=-2\\\\b=5

So the Linear System would be the shown below:

\left \{ {{y=-2x+5} \atop {y=-2x+5}} \right.

3 0
3 years ago
Calculate the mean 1,9,10,12,13,13
ExtremeBDS [4]

Answer:

By adding all the numbers and then divding them by 6 (since thats how many numbers there are) you will get 9.833...

5 0
3 years ago
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A painting company will paint this wall of a building. The owner gives him the following dimensions.
Masja [62]

The given area of the 52\frac{1}{2} ft. by 33 ft. wall less the areas of the Windows and the Door give the area of the painted part of the wall as  \underline{1642\frac{9}{16}} ft.²

<h3>How can the area of the painted part be calculated?</h3>

The area of the painted part of the wall is \underline{1642\frac{9}{16}} ft.²

The given dimensions of the wall are;

The width of the wall = 52\frac{1}{2} ft.

The height of the wall = 33 ft.

The painted area = Area of the wall - (Area of the widows + The door)

Which gives;

52\frac{1}{2} \times 33 - \left(6\frac{1}{4} \times 5\frac{3}{4} + 3\frac{1}{8} \times 4 + 9\frac{1}{2}  + 4 \times 8 \right) = \mathbf{ 1642 \frac{9}{16}}

The area of the painted part of the wall is \underline{1642\frac{9}{16}} ft.²

Learn more about the area of a rectangle here:

brainly.com/question/18101587

7 0
3 years ago
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