Write an equation that expresses the following relationship.

valentinak56 [21]

So each of these statements are talking about the square footage of land per person. Let's go and find it!

First off, let's find the number in each building of the original complex:
280 people / 4 buildings

Each building has an equal number of residents. So just divide:
280/4 = 70.

So 70 residents per building

Now consider the fact that once a new building is built, another 70 people will move in.
280 + 70= 350

350 people total.

Then lets look at the plot of land
Originally, there are 200,000 square feet of land for the 4 buildings. Then after the expansion, the plot of land will be:
200,000 + 200*200
= 200,400

Go back to the question. What's the effect of the expansion in terms of square feet of land per person?
Divide!
200,400 / 350
Approximately = 572.57 square feet

Then since it's being compared to the amount each resident had before the expansion, do the same thing with the corresponding numbers:
200,000 / 280
Approximately = 714.29

So how much will each person's land decrease?
714.29 - 572.57 approximately = 141.72 square feet.

The answer is the first choice!

Hope this helps

4 0

3 years ago

Find the value of x in this figure.

AleksAgata [21]

Step-by-step explanation:

angle x is equal to 65

angles of triangle is equal to 180

3 0

3 years ago

Read 2 more answers

Pls help its due today will give brainliest

pentagon [3]

Answer:
A: 70 ft
B: 785 ft
C: 125 ft
D: 87.2 ft/min

Explanation:
Have done this before +

5 0

3 years ago

At age 7, Maui competed in the high jump competition and managed a personal best of 3ft. At age 11 she was jumping 5ft. Her coac

Helen [10]

The high-jumper's centre of mass is about two-thirds of the way up his body when he is standing or running in towards the take off point. He needs to increase his launch speed to the highest possible by building up his strength and speed, and then use his energy and gymnastic skill to raise his centre of gravity by H, which is the maximum that the formula U2=2gH will allow. Of course there is a bit more to it in practice! When a high jumper runs in to launch himself upwards he will only be able to transfer a small fraction of his best possible horizontal sprinting speed into his upward launch speed. He has only a small space for his approach run and must turn around in order to take off with his back facing the bar. The pole vaulter is able to do much better. He has a long straight run down the runway and, despite carrying a long pole, the world's best vaulters can achieve speeds of close to 10 metres per second at launch. The elastic fibre glass pole enables them to turn the energy of their horizontal motion 12MU2 into vertical motion much more efficiently than the high jumper. Vaulters launch themselves vertically upwards and perform all the impressive gymnastics necessary to curl themselves in an inverted U-shape over the bar,sending their centre of gravity as far below it as possible.

Pole vaulter

Let's see if we can get a rough estimate of how well we might expect them to do. Suppose they manage to transfer all their horizontal running kinetic energy of 12MU2 into vertical potential energy of MgH then they will raise their centre of mass a height of:

H=U22g

If the Olympic champion can reach 9 ms−1 launch speed then since the acceleration due to gravity is g=10 ms−2 we expect him to be able to raise his centre of gravity height of H=4 metres. If he started with his centre of gravity about 1.5 metres above the ground and made it pass 0.5 metres below the bar then he would be expected to clear a bar height of 1.5+4+0.5=6 metres. In fact, the American champion Tim Mack won the Athens Olympic Gold medal with a vault of 5.95 metres (or 19′614" in feet and inches) and had three very close failures at 6 metres, knowing he had already won the Gold Medal, so our very simple estimates turn out to be surprisingly accurate.

John D. Barrow is Professor of Mathematical Sciences and Director of the Millennium Mathematics Project at Cambridge University.

3 0

3 years ago

A lake currently has a depth of 30 meters. As sediment builds up in the lake, its depth decreases by 2% per year. This situation

lubasha [3.4K]

Answer:

This situations represents the depth of the lake after t years.

The rate of decay is equal to 0.02 a year.

So the depth of the lake each year is 0.98 times the depth in the previous year.

It will take 5.77 years for the depth of the lake to reach 26.7 meters.

Step-by-step explanation:

Exponential equation of decay: