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KonstantinChe [14]

3 years ago

6

Suppose I want to build a garden abutting my house, using the house as the west side of the garden. The south and east sides of

the garden will face the street, so there I will use a decorative wall costing $30 per foot. The north side of the garden will use a tasteful fence, costing $20 per foot. If I want a 540 square foot garden, what is the minimum it will cost me?

1 answer:

Kruka [31]3 years ago

8 0

<h2>Minimum cost is 1800 $</h2>

Step-by-step explanation:

Let l be the length of west side and w be the length of north side.

East side = West side = l

North side = South side = w

Area = lw = 540 ft²

We have

$lw=540\\\\w=\frac{540}{l}$

Cost for fencing

C = 30 w + 30 l + 20 w = 30 l + 50 w

$C=30l+50\times \frac{540}{l}\\\\C=30l+\frac{27000}{l}$

At minimum cost derivative is zero,

That is

$dC=30dl-\frac{27000}{l^2}dl\\\\30-\frac{27000}{l^2}=0\\\\30=\frac{27000}{l^2}\\\\l^2=900\\\\l=30ft$

We have

lw = 540

30 x w =540

w = 18 ft

Cost = 30 l + 50 w =30 x 30 + 50 x 18 = 1800 $

Minimum cost is 1800 $

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Height versus age domain and range

lawyer [7]

Answer:

While there is a strong relationship between the two, it would certainly be ridiculous to talk about a tree with a circumference of –3 feet, or a height of 3000 feet. When we identify limitations on the inputs and outputs of a function, we are determining the domain and range of the function.

Step-by-step explanation:

3 0

3 years ago

An airplane makes a 400-mile trip against a head wind in 4 hours. The return trip takes 2.5 hours, the wind now being a tall win

tangare [24]

Answer:

$v_p=130\ miles/h$

$v_w=30\ miles/h$

Step-by-step explanation:

Lets call $v_p$ the the still-air speed of the plane and $v_w$ the speed of the wind. On the first part the speed of the plane relative to the ground will be $v_1=v_p-v_w$ , and on the second part it will be $v_2=v_p+v_w$ .

We know that the distance d=400 miles on the first and second parts are the same, so by the definition of velocity we will have:

$v_1=\frac{d}{t1}$

$v_2=\frac{d}{t2}$

Or:

$v_p-v_w=\frac{d}{t1}$

$v_p+v_w=\frac{d}{t2}$

Adding both equations:

$(v_p-v_w)+(v_p+v_w)=2v_p=\frac{d}{t1}+\frac{d}{t2}$

$v_p=\frac{1}{2}(\frac{d}{t1}+\frac{d}{t2})$

Which for our values is:

$v_p=\frac{1}{2}(\frac{400\ miles}{4h}+\frac{400\ miles}{2.5h})=130\ miles/h$

If instead of adding, we substact, we would have:

$(v_p+v_w)-(v_p-v_w)=2v_w=\frac{d}{t2}-\frac{d}{t1}$

$v_w=\frac{1}{2}(\frac{d}{t2}-\frac{d}{t1})$

Which for our values is:

$v_w=\frac{1}{2}(\frac{400\ miles}{2.5h}-\frac{400\ miles}{4h})=30\ miles/h$

6 0

4 years ago

use the graph that shows the averge number of the heartbeats for an active adult brown bear and hibernating brown bear

julia-pushkina [17]

<span>Given the graph that shows the average number of the heartbeats for an active adult brown bear and hibernating brown bear

Part a</span>.
What does the point (2, 120) represent on the graph?

From the graph the x axis represent the time in minutes while the y axis represent the number of heartbeats.

Therefore, point (2, 120) means that it takes an active adult brown bear an average of 2 minutes to have 120 heartbeat.

Part b.
What does the ratio of the y-coordinate to the x-coordinate for each pair of points on the graph represent?

From the graph the x axis represent the time in minutes while the y axis represent the number of heartbeats.

Therefore, the ratio of the y-coordinate to the x-coordinate for each pair of points on the graph represents the average number of heatbeats of an active brown dear and a hibernating brown dear per minute.

Part c.
Use the graph to find the bear’s average heart rate when it is active and when it is hibernating.

The bear's average heart rate is given by the ratio of the y-coordinate to the x-coordinate for each pair of points on the graph.

This can be obtained by taking any two points from the graph and taking the slope.

Recall that the slope of a line is given by
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For the average hearbeat of an active brown bear, using the points (1.5, 90) and (2, 120), we calculate the average heat rate as follows:
Average heart rate = $\frac{120-90}{2-1.5} = \frac{30}{0.5} =60 \ beats \ per \ minute$

Also, for the average hearbeat of a hibernating brown bear, using the points (1.5, 18) and (2, 24), we calculate the average heat rate as follows:
Average heart rate = $\frac{24-18}{2-1.5} = \frac{6}{0.5} =12 \ beats \ per \ minute$

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2 years ago

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The answer is 8 because 7.96 ends in a number over 5 making the 9 and 10 and increasing 7 to 8

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3 years ago

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