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3 years ago

The surface area of a typical classroom floor is closest to______.

2 answers:

4 0

Answer would be C. Why A would be incorrect: 100m^2 is 100m × 100 m. That would be as big as a football field. Why B would be incorrect: 1cm^2 is 1cm × 1cm. You can use a regular 15 cm to measure a piece of paper with the length of 1cm each side. Why C would be correct: 1m^2 = 1m × 1m and it is also equals to 60cm × 60 cm. A 1 meter ruler = 60 cm = 4 times of your average 15 cm ruler. So, it is reasonable that a classroom is 10m by 10m in length and breadth. Why D would be incorrect: The same explanation applies to here as well. Things that can be measured with 1m × 1m is a square table so your classroom can't be that small to fit an average of 30 to 40 students in there. I hope my explanations were detailed and easy to understand. :)

Trava [24]3 years ago

3 0

The correct answer to this question is letter "C. 10 m^2." The surface area of a typical classroom floor is closest to 10 meters squared. Choice A is too big for a classroom. Choice B is very very small to be used as a classroom. Choice D is quite not that enough.

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Which of the following is a solution of the equation -5x + 2 = -13: answers: x=-9, x=75 x=3 x=-3

slamgirl [31]

Answer:

x= 3

Step-by-step explanation:

1. -5(3) + 2 = -13

2. -15 + 2 = -13 A positive times a negative will always equal a negative

3. -13 = -13 Whenever you add anything to a negative number, the

number becomes smaller

Ex: -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, ...........

Although the number itself becomes smaller, its value

becomes bigger

Hope this helps.

7 0

3 years ago

Mr. Fink's economy car can travel 420 miles on a 12-gallon tank of gas. Determine how many miles he can travel on 8 gallons.

Sphinxa [80]

Answer:

280 miles

Step-by-step explanation:

So, 420 miles=12 gallons

That means that for 1 gallon you would be able to go 35 miles.

Just multiply that by 8. 35*8=280

3 0

3 years ago

How may pound is 1 ton

SVETLANKA909090 [29]

2,000 IBs (pounds) are in a ton.

3 0

3 years ago

Read 2 more answers

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago

Select the postulate that specifies the minimum number of points in space.

rusak2 [61]

I think step five is correct but no sure

5 0

3 years ago

Read 2 more answers