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Shalnov [3]
3 years ago
11

Please help me answer this question

Mathematics
1 answer:
SOVA2 [1]3 years ago
8 0
Well knowing that it’s 5 1/4 and the cost of paint is 13.99, I’m going to assume to multiply by the given product which is 13.99 by one quarter (which is 25)
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Zoe jogs 3/4 of a mile every 1/5 of an hour.
stepan [7]

Answer:

15/4 please let me know if this helped you :)

Step-by-step explanation:

0.75*5=3.75 or 15/4

all of your answers are less than 3/4ths of a mile...she is increasing the amount she runs not decreasing.

3 0
2 years ago
Find the diameter and radius of a circle with a circumference of 16 pi in
ELEN [110]
16 pi?  well... 16 pi = 50.2654824574. So... the diameter would be 16 and the radius would be 8. I think!
4 0
2 years ago
Read 2 more answers
please show on graph (with x and y coordinates) state where the function x^4-36x^2 is non-negative, increasing, concave up​
babunello [35]

Answer:

y'' =12x^2 -72=0

And solving we got:

x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}

We can find the sings of the second derivate on the following intervals:

(-\infty Concave up

x=-\sqrt{6}, y =-180 inflection point

(-\sqrt{6} Concave down

x=\sqrt{6}, y=-180 inflection point

(\sqrt{6} Concave up

Step-by-step explanation:

For this case we have the following function:

y= x^4 -36x^2

We can find the first derivate and we got:

y' = 4x^3 -72x

In order to find the concavity we can find the second derivate and we got:

y'' = 12x^2 -72

We can set up this derivate equal to 0 and we got:

y'' =12x^2 -72=0

And solving we got:

x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}

We can find the sings of the second derivate on the following intervals:

(-\infty Concave up

x=-\sqrt{6}, y =-180 inflection point

(-\sqrt{6} Concave down

x=\sqrt{6}, y=-180 inflection point

(\sqrt{6} Concave up

8 0
2 years ago
What is the scale factor of the dilation? Help please
DaniilM [7]
Imagine A'V' is part of the dilated figure, and AV is part of the original figure.

A'V' = 3.2

AV = 3.2 + 4.8 = 8

A'V'/AV = 3.2/8 = 2/5

Answer: First choice, 2/5.
4 0
3 years ago
Can someone put an example of how to do bearings for precal?
Reika [66]

Answer:

A bearing is used to represent the direction of one point relative to another point. For                                                                                                                          example, the bearing of A from B is 065º. The bearing of B from A is 245º.

4 0
2 years ago
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