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4 years ago

Someone help please!!!

Mathematics

1 answer:

Ne4ueva [31]4 years ago

6 0

Answer:

41.5 ft

Step-by-step explanation:

From the information given, and definition of midpoint, we know that:

AB = BC = ½ AC

AC = CE = ½ AE

CD = DE = ½ CE

GF = FE = ½ GE

HG = GE = ½ HE

AI = IH = ½ AH

AJ = JI = ½ AI

We also know:

AH = 20

HE = 14

GD = 4

Therefore:

AI = IH = 10

HG = GE = 7

AJ = JI = 5

GF = FE = 7/2

Next, since JB and IC are parallel with HE, we know that AJB and AIC are similar to AEH. So:

JB / 5 = 14 / 20

JB = 7/2

IC / 10 = 14 / 20

IC = 7

And since DF and CG are parallel to AH, then DFE and DGE are similar to AHE. So:

DF / (7/2) = 20 / 14

DF = 5

CG / 7 = 20 / 14

CG = 10

Next we know that AIB and AHC are similar, and DEG and CEH are similar.

IB / 10 = CH / 20

CH / 14 = 4 / 7

CH = 8, IB = 4.

We've found all the lengths inside triangle AEH. Adding them up:

JB + IB + IC + CH + CG + DG + DF

7/2 + 4 + 7 + 8 + 10 + 4 + 5

41.5

The total length of the inside bars is 41.5 ft.

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find the slope of the curve y=x^2-2x-5 at the point P(2,5) by finding the limit of secant slopes through point P

Fynjy0 [20]

The point (2, 5) is not on the curve; probably you meant to say (2, -5)?

Consider an arbitrary point Q on the curve to the right of P, $(t,y(t))=(t,t^2-2t-5)$ , where $t>2$ . The slope of the secant line through P and Q is given by the difference quotient,

$\dfrac{(t^2-2t-5)-(-5)}{t-2}=\dfrac{t^2-2t}{t-2}=\dfrac{t(t-2)}{t-2}=t$

where we are allowed to simplify because $t\neq2$ .

Then the equation of the secant line is

$y-(-5)=t(x-2)\implies y=t(x-2)-5$

Taking the limit as $t\to2$ , we have

$\displaystyle\lim_{t\to2}t(x-2)-5=2(x-2)-5=2x-9$

so the slope of the line tangent to the curve at P as slope 2.

- - -

We can verify this with differentiation. Taking the derivative, we get

$\dfrac{\mathrm dy}{\mathrm dx}=2x-2$

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4 0

3 years ago

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Alex777 [14]

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1 year ago

When chad had one cat he needed to serve 1/5 of a can of cat food each day. Now that chad adopted a second cat he needs to serve

Akimi4 [234]

Answer:

$5\frac{3}{10}$ cans extra food is needed to feed the second cat.

Step-by-step explanation:

Given:

When chad had one cat he needed to serve 1/5 of a can of cat food each day. Now that chad adopted a second cat he needs to serve a total of 11/2 cans each day.

Now, to find the extra food is needed to feed the second cat.

Food needed for one cat = $\frac{1}{5}\ of\ can.$