SOCIAL STUDIES HIGH SCHOOL

West of a city, a certain eastbound route is straight and makes a steep descent toward the city. The highway has a 11% grade, which means that its slope is − 11 100 . Driving on this road, you notice from elevation signs that you have descended a distance of 1000 ft. What is the change in your horizontal distance in miles?

Answers

Answer 1

Answer:

The change in the horizontal distance would be as follows:

$1.721$ miles

Find the distance

Given that,

Grade of the highway $= 11$ %

Distance for every horizontal descent measuring 11 ft $= 100 ft.$

Distance for every horizontal descent measuring 1 ft $= 100/11$

Distance for every horizontal descent measuring 1000 ft $= 1000$ × $100/11$

$= 9090.90 ft$

Since,

$1 mile = 5280 ft$

$1 ft. = 0.000189394 miles$

So,

∵ $9090.90 ft$ $= 9090.90 ft$ × $0.000189394 miles$

$= 1.721 miles$

Thus, $1.721$ miles is the correct answer.

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Answer 2

Answer:

Answer:

1.721 miles

Explanation:

Given:

Grade = 11%

i.e for every 11 ft of descend horizontal distance is 100 ft

for every 1 ft of descend horizontal distance is $\frac{\textup{100}}{\textup{11}}$ ft

therefore,

For the descend of 1000 ft horizontal distance = $1000*\frac{\textup{100}}{\textup{11}}$ ft

= 9090.90 ft

Also,

1 mile = 5280 ft

1 ft = 0.000189394 miles

therefore,

9090.90 ft = 9090.90 ft × 0.000189394 miles

= 1.721 miles

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Answers

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Hope I helped.

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What has been a constant criticism of China and its national practices?a. weak military with no effective leadership
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Answers

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David is choosing between two exercise routines. In Routine #1, he burns calories walking. He then runs at a rate that burns calories per minute. In Routine #2, he burns calories walking. He then runs at a rate that burns calories per minute. For what amounts of time spent running will Routine #1 burn at most as many calories as Routine #2? Use for the number of minutes spent running, and solve your inequality for .

Answers

Answer:

The answer is below

Explanation:

The question is not complete, the complete question is in the form of: David is choosing between two exercise routines. In Routine #1, he burns 20 calories walking. He then runs at a rate that burns 10.5 calories per minute. In Routine #2, he burns calories 38 walking. He then runs at a rate that burns 8.5 calories per minute. For what amounts of time spent running will Routine #1 burn at most as many calories as Routine #2? Use for the number of minutes spent running, and solve your inequality for .

Answer:

Let us assume that the number of minutes spent running is t minute. The equation that represents the total calories burnt for routine 1 is given as:

20 + 10.5t

While the total calories burnt for routine 2 is given as:

38 + 8.5t

Since Routine #1 burn at most as many calories as Routine #2, hence it can be represented by the inequality

20 + 10.5t < 38 + 8.5t

Solving the inequality:

10.5t - 8.5t < 38 - 20

2t < 18

t < 9 minutes

For routine 1 to burn at most as many calories as routine 2, the time spent running must be less than 9 minutes

Final answer:

David should run for less than or equal to (W2 - W1) / (R1 - R2) minutes for the calories burned in Routine #1 to be at most equal to that of Routine #2. W1, W2, R1, and R2 represent the number of calories burned walking and running rate in each routine, respectively.

Explanation:

The question doesn't provide specific figures for the amount of calories burned through walking or running in either routine. Therefore, for our purposes, let's denote the number of calories burned walking in Routine #1 and #2 as W1 and W2 respectively, and the rate of calories burned per minute running as R1 and R2 respectively.

If x represents the time (in minutes) David spends running, the total number of calories burned in Routine #1 would be W1 + R1*x, and for Routine #2 it would be W2 + R2*x.

David would burn at most as many calories with Routine #1 as he would with Routine #2 when W1 + R1*x ≤ W2 + R2*x. To solve this inequality for x, you would perform the following steps:

Subtract W1 from both sides of the inequality to isolate the variable terms on one side: R1*x ≤ W2 - W1 + R2*x
Subtract R2*x from both sides: (R1 - R2) * x ≤ W2 - W1
Divide by R1 - R2 to solve for x: x ≤ (W2 - W1) / (R1 - R2)
The amount of time spent running for which Routine #1 burns at most as many calories as Routine #2 is given by x ≤ (W2 - W1) / (R1 - R2).

Learn more about Inequalities here:

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Answers

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Answers

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