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3 years ago

How is it 86 I neeeed Explainatiob pleaseeee!!!!!

Mathematics

1 answer:

givi [52]3 years ago

7 0

Answer:

it is 86

Step-by-step explanation:

House 1 has 6 sticks. House 2 has 11, house 3 has 16, and the pattern goes on. Each following house uses 5 more sticks. At design 17, we can use this pattern to solve for it. We would do 5*17, remembering to add 1 as house 1 has its own additional stick. 5*17 + 1 = 86

Hope this helps, let me know if you have any questions :))

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Find the local maximum and minimum values of f using both the first and second derivative tests. f(x) = 6 9x2 − 6x3 local maximu

olga55 [171]

The local minimum value of the function f(x) = 6 + 9x² - 6x³ is 6 and the local maximum value of the function f(x) = 6 + 9x² - 6x³ is 9.

For given question,

We have been given a function f(x) = 6 + 9x² - 6x³

We need to find the local maximum and local minimum of the function f(x)

First we find the first derivative of the function.

⇒ f'(x) = 0 + 18x - 18x²

⇒ f'(x) = - 18x² + 18x

Putting the first derivative of the function equal to zero, we get

⇒ f'(x) = 0

⇒ - 18x² + 18x = 0

⇒ 18(-x² + x) = 0

⇒ x (-x + 1) = 0

⇒ x = 0 or -x + 1 = 0

⇒ x = 0 or x = 1

Now we find the second derivative of the function.

⇒ f"(x) = - 36x + 18

At x = 0 the value of second derivative of function f(x),

⇒ f"(0) = - 36(0) + 18

⇒ f"(0) = 0 + 18

⇒ f"(0) = 18

Here, at x=0, f"(x) > 0

This means, the function f(x) has the local minimum value at x = 0, which is given by

⇒ f(0) = 6 + 9(0)² - 6(0)³

⇒ f(0) = 6 + 0 - 0

⇒ f(0) = 6

At x = 1 the value of second derivative of function f(x),

⇒ f"(1) = - 36(1) + 18

⇒ f"(1) = - 18

Here, at x = 1, f"(x) < 0

This means, the function f(x) has the local maximum value at x = 1, which is given by

⇒ f(1) = 6 + 9(1)² - 6(1)³

⇒ f(1) = 6 + 9 - 6

⇒ f(1) = 9

So, the function f(x) = 6 + 9x² - 6x³ has local minimum at x = 0 and local maximum at x = 1.

Therefore, the local minimum value of the function f(x) = 6 + 9x² - 6x³ is 6 and the local maximum value of the function f(x) = 6 + 9x² - 6x³ is 9.

Learn more about the local minimum value and local maximum value here:

brainly.com/question/15437371

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3 0

2 years ago

An exam worth 388 points contains 65 questions. Some questions are worth 4 points and the others are worth 8 points. How many 4

BaLLatris [955]

Answer:

33 questions.

Step-by-step explanation:

Let x = 4 points questions.

Let y = 8 points questions.

Translating the word problem into an algebraic equation;

For the total number of questions;

$x + y = 65$ .........equation 1

For the total number of points;

$4x + 8y = 388$ ..........equation 2

Given the following data;

Total number of questions = 65

Total number of points = 333

Solving the linear equation by using the substitution method;

Making x the subject in equation 1:

$x = 65 - y$ .......equation 3

Substituting "x" into equation 2;

$4(65 - y) + 8y = 388$

Simplifying the equation, we have;

$260 - 4y + 8y = 388$

$260 + 4y = 388$

Rearranging the equation, we have;

$4y = 388 - 260$

$4y = 128$

$y = \frac {128}{4}$

y = 32 (For the 8 points question).

Substituting "y" into equation 3;

$x = 65 - 32$

x = 33 (For the 4 points question).

Therefore, the number of 4 point questions on the test are 33 questions.

8 0

3 years ago

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$