3

Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types of young animals. The thre

andrew-mc [135]

Answer:

The area function is

$A=\frac{135}{2}x-\frac{9}{2}x^2$ .

The domain and range of A is $(0,15m)$ and $(0, 253.125 m^2]$ .

Step-by-step explanation:

The given length of fencing is $90 m$ .

Let the length and width of each pen be $x$ and $y$ respectively as shown in the figure.

As there are 3 pens, so, the total area,

$A= 3 xy \;\cdots (i)$

From the figure the total length of fencing is $6x+4y$ .

Here, for a significant area for the animals, $x>0$ as well as $y>0$ as $x$ and $y$ are the sides of ben.

From the given value:

$6x+4y=90\;\cdots (ii)$

$\Rightarrow y=\frac {45}{2}-\frac{3x}{2}$

Now, from equation (i)

$A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)$

$\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)$

This is the required area function in the terms of variable $x$ .

For the domain of area function, from equation (ii)

$x=15-\frac{2y}{3}$

$\Rightarrow x$ [as y>0]

So, the domain of area function is $(0,15m)$ .

For the range of area function:

As $x \rightarrow 0$ or $y\rightarrow 0$ , then $A\rightarrow 0$ [from equation (i)]

$\Rightarrow A>0$

Now, differentiate the area function with respect to $x$ .

$\frac {dA}{dx}=\frac{135}{2}-9x$

Equate $\frac {dA}{dx}$ to zero to get the extremum point.

$\frac {dA}{dx}=0$

$\Rightarrow \frac{135}{2}-9x=0$

$\Rightarrow x=\frac{15}{2}$

Check this point by double differentiation

$\frac {d^2A}{dx^2}=-9$

As, $\frac {d^2A}{dx^2}$ , so, point $x=\frac{15}{2}$ is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

$A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2$

$\Rightarrow A=253.125 m^2$

Hence, the range of area function is $(0, 253.125 m^2]$ .

4 0

4 years ago

Evaluate the function. f(x) = 2x2 Find f(-3)

Ede4ka [16]

F(-3)=-12
because f(x) is y=2*2=4

4 0

3 years ago

Read 2 more answers

Recall the question for a circle with center. (H,k) and Radius r. At what point in the first quadrant does the line with equatio

kap26 [50]

Answer:

the point lies at (1.7, 3.6) in the first quadrant.

Step-by-step explanation:

The formula for a circle with centre (h,k) and radius (r) can be expressed as :

(x-h)² + (y-k)² = r²

Now for the expression of the circle with radius 3 and center (0,1), we have:

(x - 0)² + (y - 1)² = 3²

x² + (y-1)² = 9

replacing y = 1.5x + 1 in the above equation, we have:

x² + (( 1.5x + 1 ) - 1 )² = 9

x² + ( 1.5x +1 - 1)² = 9

x² + (1.5x)² = 9

x² + 2.25x² = 9

3.25x² = 9

x² = 9/3.25

$x^2 = \dfrac{9}{\dfrac{13}{4}}$

$x^2 = {9} \times {\dfrac{4}{13}}$

$x= \sqrt{{9} \times {\dfrac{4}{13}}}$

$x= \pm 3 \times \dfrac{2}{\sqrt{13}}}$

$x= \pm \dfrac{6}{\sqrt{13}}}$

x = 1.7 in the positive x - axis

Recall that:

y = 1.5x + 1

y = 1.5 (1.7) +1

y = 2.55 +1

y = 3.55

y = 3.6

Therefore, the point lies at (1.7, 3.6) in the first quadrant.

4 0

3 years ago

Find the area of the following shape. Show all work

Pavel [41]

Best way to solve this is by using

$\sqrt{s(s - a)(s - b)(s - c)}$

$where \: s = \frac{a + b + c}{2}$

s=(12+8+17)/2

=18.5

using the formulae

area =43.5

4 0

3 years ago

Explain why the equation 5(2x+1)-2=6x+5 has only one solution. Then find the solution.

netineya [11]

Answer:

10x + 3 = 6x + 5

Based on this, we have the lines y=10x+3 and y=6x+5. Both of the equations are linear, so they are straight lines when graphed. A system of straight lines can only have one solution maximum.

Solution:

x = 1/2

y = 8

(1/2 , 8)

Step-by-step explanation:

Start by expanding with the distributive property, then simplify by collecting like terms.

5(2x + 1) - 2 = 6x + 5 Multiply "5" with each term in the brackets.

10x + 5 - 2 = 6x + 5 Collect left side's like terms (5 - 2 = 3)

10x + 3 = 6x + 5

Based on this, we have the lines y=10x+3 and y=6x+5. A solution is the same as the POI (point of intersection). Both of the equations are linear, so they are straight lines when graphed. A system of straight lines can only have one solution maximum.

Some linear systems might have no solutions (when the lines are parallel) or infinite solutions (when the lines are equivalent, the same).

To find the solution, continue with the equation and isolate "x". This will give you the x-coordinate of the point of intersection.

To isolate 'x', move everything to the other side of 'x'. To move something, you do its reverse operation to both sides in reverse BEDMAS order.

10x + 3 = 6x + 5

10x - 6x + 3 = 6x - 6x + 5 Subtract 6x from both sides

4x + 3 = 5 "6x" cancelled out on the right. Simplified left (10x-6x=4x)

4x + 3 - 3 = 5 - 3 Subtract 3 on both sides. "3" on the left will cancel out

4x = 2 The right side was simplified (5 - 3 = 2).

4x/4 = 2/4 Divide both sides by 4 to isolate 'x'. Simplify right side.

x = 1/2 Solution's x-coordinate

Substitute what you found for 'x' into any equation that has 'y' (either y=10x+3 or y=6x+5). I will choose the first equation.

y = 10x + 3

y = 10(1/2) + 3 Multiply 1/2 and 10

y = 5 + 3 Add

y = 8 Solution's y-coordinate

Therefore the solution is when x=1/2 and y=8, or as an ordered pair, (1/2 , 8).

8 0

3 years ago