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3 years ago

13

H(x) = x^2 - 1 Over which interval does h have a negative average rate of change?

1 answer:

vredina [299]3 years ago

6 0

The Curve has negative rate of change for negative values of x

I.e in the interval - infinity to 0

Step-by-step explanation:

Step 1:

The change in the rate for the given function is obtained calculating the first derivative.

So dy/dx = dH(x)/dx = 2*x

Step 2:

This rate of change will be negative when the dy/ dx is less than 0

So dy/dx = 2x < 0 => x< 0

So for values of x less than 0 the function has negative values for the rate of change.

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Jake has $4,000 invested at 8% annual interest. He has $2,800 more to invest. At what rate

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3 years ago

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Answer:

The area of rectangle BEFD is 180 square units.

Step-by-step explanation:

After checking the figure given, we have the following information:

$AD = 15$ , $AB = 12$ , $BC = 15$ , $CD = 12$

By Pythagorean Theorem, we determine the length of the line segment BD:

$BD = \sqrt{AB^{2}+AD^{2}}$ (1)

$BD = \sqrt{12^{2}+15^{2}}$

$BD = 3\sqrt{41}$

In addition, we know the following characteristics of the rectangle BEFD:

$BD = EF$ , $EF = EC + CF$ , $BE = FD$ (2), (3), (4)

By Pythagorean Theorem:

$BC^{2} = BE^{2}+EC^{2}$ (5)

$CD^{2} = CF^{2}+DF^{2}$ (6)

By (3), (4), (5) and (6):

$BC^{2} = BE^{2} + EC^{2}$ (7)

$CD^{2} = (EF-EC)^{2} + BE^{2}$ (8)

By (7) in (8):

$CD^{2} = (EF-EC)^{2}+ (BC^{2}-EC^{2})$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC + EC^{2}+BC^{2}-EC^{2}$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC +BC^{2}$

Then, we clear $EC$ :

$2\cdot EF\cdot EC = EF^{2} + BC^{2} - CD^{2}$

$EC = \frac{EF^{2}+BC^{2}-CD^{2}}{2\cdot EF}$

If we know that $EF = 3\sqrt{41}$ , $BC = 15$ and $CD = 12$ , then the length of the segment $EC$ is:

$EC = \frac{75\sqrt{41}}{41}$

And the length of the line segment $CF$ is:

$CF = EF - EC$

$CF = 3\sqrt{41}-\frac{75\sqrt{41}}{41}$

$CF = \frac{48\sqrt{41}}{41}$

And the length of the line segment $DF$ is determined by Pythagorean Theorem:

$FD = \sqrt{CD^{2}-CF^{2}}$

$FD = \sqrt{12^{2}-\left(\frac{48\sqrt{41}}{41} \right)^{2}}$

$FD = \frac{60\sqrt{41}}{41}$

And the area of the rectangle is determined by the following formula:

$A = FD\cdot EF$

$A = \left(\frac{60\sqrt{41}}{41} \right)\cdot (3\sqrt{41})$

$A = 180$

The area of rectangle BEFD is 180 square units.

4 0

3 years ago