9x + 8y3 – 7 = 3y3 -3<br> 9x+ 5y = 0

If you are offered one slice from a round pizza (in other words, a sector of a circle) and the slice must have a perimeter of 24

VARVARA [1.3K]

Answer:

A 12 in diameter will reward you with the largest slice of pizza.

Step-by-step explanation:

Let $r$ be the radius and $\theta$ be the angle of a circle.

According with the graph, the area of the sector is given by

$A=\frac{1}{2}r^2\theta$

The arc lenght of a circle with radius $r$ and angle $\theta$ is $r \theta$

The perimeter of the pizza slice is composed of two straight pieces, each of length r inches, and an arc of the circle which you know has length s = rθ inches. Thus the perimeter has length

$2r+r\theta=24 \:in$

We need to express the area as a function of one variable, to do this we use the above equation and we solve for $\theta$

$2r+r\theta=24\\r\theta=24-2r\\\theta=\frac{24-2r}{r}$

Next, we substitute this equation into the area equation

$A=\frac{1}{2}r^2(\frac{24-2r}{r})\\A=\frac{1}{2}r(24-2r)\\A=12r-r^2$

The domain of the area is

$0$

To find the diameter of pizza that will reward you with the largest slice you need to find the derivative of the area and set it equal to zero to find the critical points.

$\frac{d}{dr} A=\frac{d}{dr}(12r-r^2)\\A'(r)=\frac{d}{dr}(12r)-\frac{d}{dr}(r^2)\\A'(r)=12-2r$

$12-2r=0\\-2r=-12\\\frac{-2r}{-2}=\frac{-12}{-2}\\r=6$

To check if $r=6$ is a maximum we use the Second Derivative test

if $f'(c)=0$ and $f''(c)$ , then f(x) has a local maximum at x = c.

The second derivative is

$\frac{d}{dr} A'(r)=\frac{d}{dr} (12-2r)\\A''(r)=-2$

Because $A''(r)=-2$ the largest slice is when r = 6 in.

The diameter of the pizza is given by

$D=2r=2\cdot 6=12 \:in$

A 12 in diameter will reward you with the largest slice of pizza.

3 0

2 years ago

Giving brainliest and max rating.

Yanka [14]

Given:

The cost of each carnival ticket is $5.

To find:

The equation, table of values and graph for the given problem.

Solution:

Let x be the number of tickets and y be the total money spent on tickets.

Cost of one ticket = $5

Cost of x tickets = $5x

So, total cost is

$y=5x$

The required equation is $y=5x$ .

At x=1,

$y=5(1)$

$y=5$

At x=2,

$y=5(2)$

$y=10$

At x=3,

$y=5(3)$

$y=15$

The required table of values is

x y

1 5

2 10

3 15

So, the required table of values is table A.

From the above table, it is clear that the graph passes through the point (1,5), (2,10) and (3,15). The graph B passes through these points.

So, the required graph is graph B.

Since the required answers are $y=5x$ , table A, graph B, therefore the correct option is B.

8 0

2 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$