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tankabanditka [31]
3 years ago
11

Need help solving this problem

Mathematics
1 answer:
Digiron [165]3 years ago
7 0
2x^3+9, I just used an online calculator
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1. A function has a second rate of change of -4. Does the function have a maximum or a minimum?
Aloiza [94]

This function would have a maximum.

Since we are subtracting by a -4 for each increase in x, we know that the numbers will continue to go down. Given this fact, we know the number will never be higher than when we started, but the number could go infinitely low. As a result we have a maximum and no minimum.

8 0
3 years ago
(2r^3-5r) - (r^3+5r)
kotykmax [81]

Answer:

r^3-10r

Answer Explanation:

4 0
3 years ago
Which equation is satisfied by all of the plotted points?
Lubov Fominskaja [6]
A. y=2x 
because slope-intercept form is y=mx+b. m= slope & b= y-int. The y-int is at the origin, so therefore y=0. The slope is 2, so therefore the answer is y=2x
8 0
3 years ago
1. Which form shows the slope-intercept form?<br> y = mx +b<br> Ax + By =C<br> y-y1 = m(x-x1)
KatRina [158]

Answer:

y = mx + b

Step-by-step explanation:

b is your y intercept while m is your slope which is just rise over run.

4 0
3 years ago
I am a bit lost with this problem. Could someone please explain this to me :)
Serjik [45]
They want the point D such that D is equally distant away from the three gates. Call the gates A, B, and C

They want
AD = BD = CD

If we draw a circle through those three points, then point D would be the center
This circle is known as the circumcircle as it goes through the three points
The center D is the circumcenter

So all you have to do is find the perpendicular bisector of two segments of the triangle, say of AB and BC. Then find where those perpendicular bisectors meet. That intersection would be point D. 

Since we don't have actual numbers or coordinates to work with, we can't go further and actually find out where point D is located in terms of (x,y) coordinates. However, you can still get a good idea using a compass and straightedge. 
3 0
4 years ago
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