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3 years ago

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A dog owner wants to use 200 ft of fencing to enclose the greatest possible area for his dog. he wants the fenced area to be rec

tangular. what dimensions should he use?

1 answer:

Fiesta28 [93]3 years ago

5 0

Perimeter (P) = 2 · Length(L) + 2 · Width (W) → P = 2L + 2W
Solve for either L or W (I am solving for L).
200 - 2W = 2L
(200 - 2W)/2 = L
100 - W = L

Area (A) = Length (L) · Width (W)
= (100 - W) · W
      = 100W - W²

Find the derivative, set it equal to 0, and solve:
dA/dW = 100 - 2W
   0 = 100 - 2W
   W = 50

refer to the equation above for L:
100 - W = L
100 - 50 = L
      50 = L

Dimensions for the maximum Area are 50 ft x 50 ft

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Find area of the shaded region.

Monica [59]

Answer:

The area of the shaded region is about 38.1 square centimeters.

Step-by-step explanation:

We want to find the area of the shaded region.

To do so, we can first find the area of the sector and then subtract the area of the triangle from the sector.

The given circle has a radius of 6 cm.

And the given sector has a central angle of 150°.

The area for a sector is given by the formula:

$\displaystyle A=\pi r^2\cdot \frac{\theta}{360^\circ}$

In this case, r = 6 and θ = 150°. Hence, the area of the sector is:

$\displaystyle \begin{aligned}A&=\pi(6)^2\cdot \frac{150}{360}\\ &=36\pi\cdot \frac{5}{12}\\&=3\pi \cdot 5\\&=15\pi \text{ cm}^2\end{aligned}$

Now, we can find the area of the triangle. We can use an alternative formula:

$\displaystyle A=\frac{1}{2}ab\sin(C)$

Where a and b are the side lengths, and C is the angle between them.

Both side lengths of the triangle are the radii of the circle. So, both side lengths are 6.

And the angle C is 150°. Hence, the area of the triangle is:

$\displaystyle A=\frac{1}{2}(6)(6)\sin(150)=18\sin(150)$

The area of the shaded region is equivalent to the sector minus the triangle:

$A_{\text{shaded}}=A_{\text{sector}}-A_{\text{triangle}}$

Therefore:

$A_{\text{shaded}}=15\pi -18\sin(150)$

Use a calculator:

$A_{\text{shaded}}=38.1238...\approx 38.1\text{ cm}^2$

The area of the shaded region is about 38.1 square centimeters.

6 0

3 years ago

BRAINLIEST IF CORRECT

Softa [21]

Answer:

what's the question?????????

7 0

3 years ago

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If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form

algol13

Answer:

$\frac{d^2y}{dx^2} = \frac{-4}{3}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

<u>Step 1: Define</u>

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate <em>y'</em> terms: $-6x^2=2yy'+y'$
[Algebra] Factor <em>y'</em>: $-6x^2=y'(2y+1)$
[Algebra] Isolate <em>y'</em>: $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

<u>Step 3: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute <em>y'</em>: $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

<u>Step 4: Find Slope at Given Point</u>

[Algebra] Substitute in <em>x</em> and <em>y</em>: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Multiply: $y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
[Pre-Algebra] Simplify: $y''(-1,-2) = \frac{-4}{3}$

6 0

3 years ago

How do u solve this problem

Feliz [49]

It is a 3:1 ratio of male car drivers to female truck drivers

4 0

4 years ago

50 – 5(4.3 – 1.3) ÷ 0.5

marissa [1.9K]

Answer:

(0.5, 1.3)(0.5, 1.3)

Step-by-step explanation:

Given equations are:

As we can see that the given equations are linear equations which are graphed as straight lines on graph. The solution of two equations is the point of their intersection on the graph.

We can plot the graph of both equations using any online or desktop graphing tool.

We have used "Desmos" online graphing calculator to plot the graph of two lines (Picture Attached)

We can see from the graph that the lines intersect at: (0.517, 1.267)

Rounding off both coordinates of point of intersection to nearest tenth we get

(0.5, 1.3)

Hence,

(0.5, 1.3) is the correct answer

Keywords: Linear equations, variables

4 0

3 years ago

Read 2 more answers