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lakkis [162]
2 years ago
15

8.2.AP-5

Mathematics
1 answer:
miskamm [114]2 years ago
8 0

Answer:

parallelogram

Step-by-step explanation:

Parallelograms have two pairs of parallel sides and no right angles.

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1+4=5&lt;br /&gt;<br> 2+5=12&lt;br /&gt;<br> 3+6=21&lt;br /&gt;<br> 8+11=?
san4es73 [151]
Next time make it more easyer
3 0
3 years ago
Find the closest point to y in the subspace w spanned by bold v 1v1 and bold v 2v2. yequals=left bracket start 4 by 1 matrix 1st
Dmitrij [34]

Answer:

See attached picture.

Step-by-step explanation:

6 0
3 years ago
What is the answer to this?
nekit [7.7K]

Answer:

Up

Step-by-step explanation:

Y= 1x^2 + 3x + 9

Y= Ax^2 + Bx + C

To determine which way a parabola opens just look at the coefficient (the number) of the x^2 - Always use the coefficient that is beside the x^2

So,  if the coefficient A is positive the parabola is opening upward

      if the coefficient A is negative the parabola is opening downward

7 0
2 years ago
Read 2 more answers
Please Help: Determine which ordered pair is NOT a solution of y = -5x - 4
777dan777 [17]
It's A.
All you need to do is plug in the X from the ordered pair, then solve.

y=-5x-4
y=-5(10)-4
y=-50-4
y=-54
Which is NOT -52, so the answer is indeed A
8 0
2 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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