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3 years ago

If it took 10 hours to mow 8 lawns, then at that rate, how many lawns could be mowed in 30 hours?

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

7 0

Answer:

24 Lawns

Step-by-step explanation: If it took 10 hours to mow 8 lawns, then add 8 to its self 3 times and you get 24.

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Consider the following figure.

Karolina [17]

Answer:

BD = 16

Step-by-step explanation:

By applying mean proportional theorem in the given right triangles,

$\frac{AD}{BD}=\frac{BD}{CD}$

AD × CD = BD²

8 × 32 = BD²

BD = √256

BD = 16

Therefore, measure of side BD = 16 units.

4 0

3 years ago

How would I do the steps to solve this?

allsm [11]

Answer:

The maximum revenue is 16000 dollars (at p = 40)

Step-by-step explanation:

One way to find the maximum value is derivatives. The first derivative is used to find where the slope of function will be zero.

Given function is:

$R(p) = -10p^2+800p$

Taking derivative wrt p

$\frac{d}{dp} (R(p) = \frac{d}{dp} (-10p^2+800p)\\R'(p) = -10 \frac{d}{dp} (p^2) +800 \ frac{d}{dp}(p)\\R'(p) = -10 (2p) +800(1)\\R'(p) = -20p+800\\$

Now putting R'(p) = 0

$-20p+800 = 0\\-20p = -800\\\frac{-20p}{-20} = \frac{-800}{-20}\\p = 40$

As p is is positive and the second derivative is -20, the function will have maximum value at p = 40

Putting p=40 in function

$R(40) = -10(40)^2 +800(40)\\= -10(1600) + 32000\\=-16000+32000\\=16000$

The maximum revenue is 16000 dollars (at p = 40)

3 0

3 years ago

I will mark you BRAINLIST please help with all 3 and show work on a paper please I need this thank you !!

o-na [289]

I hope you can understand this :)

8 0

3 years ago

Suppose a tub has the shape of an elliptical paraboloid given by z = ax2+by2 (where a, b are some positive constants). If a marb

Umnica [9.8K]

Answer:

It would roll in this direction.

$\nu = (-a/\sqrt{a^2+b^2},-b/\sqrt{a^2+b^2})$

Step-by-step explanation:

It would roll to the direction of maximum decrease, which is the -1 times the direction of maximum increase, which is given by the gradient of the function.

Since

$z = ax^2 + by^2$

For this case, the gradient of your function would be

$\nabla z = (2ax , 2by)$

And -1 times the gradient of your function would be

$-\nabla z = (-2ax , -2by)\\$

Then, at

$(1,1,a+b),\\x = 1 \\y = 1$

So it would go towards

$v = (-2a,-2b)$

The magnitud of that vector is

$|v| = 2\sqrt{a^2+b^2}$

and to conclude it would roll in this direction.

$\nu = (-a/\sqrt{a^2+b^2},-b/\sqrt{a^2+b^2})$

6 0

3 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago