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Ghella [55]
3 years ago
10

Which graph represents a function

Mathematics
1 answer:
artcher [175]3 years ago
3 0
Its The 3rd One Going Down I Believe 
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Determine whether the given length can be side lengths of a right triangle. 12m, 60m, 61m
Ne4ueva [31]

Answer:

No

Step-by-step explanation:

Using the converse of Pythagoras' identity.

If the longest side squared is equal to the sum of the squares of the other 2 sides then the triangle is right.

The longest side = 61, thus

61² = 3721

12² + 60² = 144 + 3600 = 3744 ≠ 3721

Thus these sides do not form a right triangle.

3 0
2 years ago
I need answers anyone help
GalinKa [24]
First one single solution second no solution 3rd one single forth one infinity and last 2 are single and infinity
7 0
3 years ago
What is the area of the figure?
son4ous [18]
Yes. 95 is correct.
You have three congruent "indentations" in the right hand side. Thus each section must be 15/3 = 5 cm long.
The top rectangle will be 8*5 = 40 cm^2
The bottom rectangle will also be 8*5 = 40 cm^2
The middle area will be 5(8-5) = 5 * 3 = 15 cm^2
40 + 40 + 15 = 95

5 0
3 years ago
What is 1/10 of 253 thousandths
Kobotan [32]
1/10×253/1000= 0.0253
5 0
3 years ago
3. (05.06 MC)
AveGali [126]

Answer:

Step-by-step explanation:

y > 3x + 10

y < -3x - 1

<h3> Part A:</h3>

You'd graph the given systems of linear inequalities the same way as you graph the linear equations.  

To graph y > 3x + 10, plot the y-intercept, (0, 10), then use the slope, m = 3 (rise 3, run 1), to plot other points on the graph.  Use a dashed line (because of the ">" symbol).  

Follow the same steps for the other linear inequality. Plot the y-intercept, (0, -1), then use the slope, m = -3 (down 3, run 1) to plot other points. Use a dashed line (because of the "<" symbol).  

Pick a test point on either side of the boundary line and plug it into the original problem.  This will help determine which side of the boundary line is the solution.  Plug in a test point that is not on the boundary line.

Use the point of origin, (0, 0) as the test point. Plug in these values into the given systems of linear inequalities to see whether it will provide a true statement.  

y > 3x + 10

0 > 3(0) + 10

0 > 0 + 10

0 > 10 (False statement).  

y < -3x - 1

0 < -3(0) - 1

0 < 0 - 1

0 < -1 (false statement).  

Since the point of origin provided a false statement to the given systems of linear inequalities, you must shade the half-plane region where it doesn't contain the test point.  

<h3>Part B: </h3>

You'll do the same process as what I've done for the test point. Plug in the values of (8, 10) into the given systems of linear inequalities. If it provides a false statement, then it means that it is not a solution to the system.  

y > 3x + 10

10 > 3(8) + 10

10 > 24 + 10

10 > 34 (False statement).  

y < -3x - 1  

10 < -3(8) - 1

10 < -24 - 1

10 < -25 (false statement).  

Therefore, (8, 10) is not a solution to the system.    

8 0
2 years ago
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