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3 years ago

15

Every 7 days a plant grows 4 centimeters.

1 answer:

masha68 [24]3 years ago

8 0

Answer:

what the problem? I dont understand

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What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 27 − x^2? show your

LUCKY_DIMON [66]

Answer:

Option B.

Step-by-step explanation:

The given curve is

$y=27-x^2$

We need to find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve $y=27-x^2$ .

Let the vertex in quadrant I be (x,y), then the vertex in quadrant II is (-x,y) .

Length of the rectangle = 2x

Width of the rectangle = y

Area of a rectangle is

$Area=Length\times width$

$Area=2x\times y$

Substitute the value of y from the given equation.

$Area=2x(27-x^2)$

$A=54x-2x^3$ .... (1)

Differentiate with respect to x.

$\frac{dA}{dx}=54-6x^2$

Equate $\frac{dA}{dx}=0$ , to find the critical points.

$0=54-6x^2$

$6x^2=54$

Divide both sides by 6.

$x^2=9$

$x=\pm 3$

The value of x can not be negative because side length can not be negative.

Substitute x=3 in equation (1).

$A=54(3)-2(3)^3$

$A=162-54$

$A=108$

The area of the largest rectangle is 108 square units.

Therefore, the correct option is B.

7 0

3 years ago

Please help me fast and show all your work

balandron [24]

Answer:

i think it will be option 3

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

Which relation is a function?

krek1111 [17]

Answer:

the top left is a function

5 0

3 years ago

I need help with finding the answer to a) and b). Thank you!

shtirl [24]

Answer:

$\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }$

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}$

$\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin(x)=-\frac{21}{29}$

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

$a=\sqrt{29^2-21^2}=20$

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

$180$

So:

$\displaystyle 90 < \frac{x}{2} < 135$

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

1)

Recall that:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}$

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 + 20/29}{2}$

Simplify:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}$

2)

Likewise:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }$

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }$

Simplify:

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}$

3)

Finally:

$\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}$

Therefore:

$\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}$

5 0

3 years ago

34.65$ leave a 15% tip how much would it be

Talja [164]

Answer:

$2.31

Step-by-step explanation:

3 0

3 years ago