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julia-pushkina [17]
3 years ago
6

If anyone knows this please help me

Mathematics
1 answer:
alexgriva [62]3 years ago
6 0

Answer:

oh hey I gotchu-

first of all you are gonna make that into two rectangle by slicing the top off if that makes sense so you now have a large rectangle (11×12) and a small one (3×8) now you use the area formula for rectangles base×height so the small one becomes (3×8)=24 and the large one is (11×12)=132. now to make them into one shape for the total area you add those two values 132+24=156 so your total area is 156m²

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Step-by-step explanation:

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The time t required to drive a certain distance varies inversely with the speed, r. If it takes 4 hours to drive the distance at
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option D

Step-by-step explanation:

4 hours = 40 miles/per hour

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3 years ago
Read the problem and examine the diagram.
yulyashka [42]

Answer:

Option A: The answer should be 10% of 6 is the correct answer.

Step-by-step explanation:

Given that:

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10% of x = 6

By solving this, we will get the value for number of students in the band.

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3 years ago
A shipping container will be used to transport several 80-kilogram crates across the
Leno4ka [110]

Answers:

  • The inequality to solve is 80x+6300 \le 26500
  • The solution to the inequality is x \le 252.5
  • Interpretation: The most crates we can load is 252 since we can only load a positive whole number of crates, and because 253 is too many.

==============================================================

Explanation:

x = number of 80-kilogram crates

x is some positive whole number

1 crate weighs 80 kilograms, so x of them will weigh 80x kilograms.

This is added on top of the 6300 kg from the other cargo already on board.

We have a total weight of 80x+6300. Let's call this T.

So T = 80x+6300

We want T to be 26500 or smaller. Otherwise, we've gone over capacity.

This must mean we want T \le 26500 which is the same as saying 80x+6300 \le 26500 after replacing T with 80x+6300.

---------------------------

Let's follow PEMDAS in reverse to isolate x

80x+6300 \le 26500\\\\80x+6300-6300 \le 26500-6300 \ \text{ ... subtract 6300 from both sides}\\\\80x \le 20200\\\\\frac{80x}{80} \le \frac{20200}{80} \ \text{ ... divide both sides by 80}\\\\x \le 252.5\\\\

The inequality sign stays the same the entire time. It would only flip if we divided both sides by a negative number.

In the last step, we get a decimal value. However, as stated earlier, x is a positive whole number.

We cannot round to 253 because that's too high. We can check x = 253 is too high by noticing that...

80x+6300 = 80*253+6300 = 26,540

which is over the 26,500 mark by 40 kg

So we must round down to the nearest whole number and find that x = 252 is the largest x value possible. Let's check to see if we're under the weight limit

80x+6300 = 80*252+6300 = 26,460

We're under the 26,500 weight limit (with 40 kg to spare)

In short, the most crates that we can load is 252 crates in addition to the other cargo that collectively weighs 6300 kg (which may or may not consist of those 80-kilogram crates).

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