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4 years ago

Whats 39511 rounded to the nearest thousand

Mathematics

2 answers:

Romashka-Z-Leto [24]4 years ago

3 0

39,511 when rounded to the nearest thousand equals too...

40,000

FrozenT [24]4 years ago

3 0

The answer is 40,000

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Find the sum of the squares of first five natural numbers.

lisov135 [29]

explanation 

1²=1. 2²=4. 3²=9. 4²=16. 5²=25

1+4+9+16+25

answer

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3 years ago

The speed of light is about nine hundred eighty million feet per second. What is that speed in scientific notation?

guajiro [1.7K]

Nine hundred eighty million is 9.8 x 10^8, which is the bottom left option.

I hope you are satisfied with my answer.

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4 years ago

Read 2 more answers

You deposit $2500 into a bank account that pays 7.5% annual interest, compounded continuously. Find

AleksandrR [38]

Answer:

16,125

Step-by-step explanation:

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3 years ago

Solve : 6(3 + n) <48.

Alecsey [184]

Answer:

n<5

Step-by-step explanation:

6(3+n)<48

Step 1: Simplify both sides of the inequality.

6n+18<48

Step 2: Subtract 18 from both sides.

6n+18−18<48−18

6n<30

Step 3: Divide both sides by 6.

6n /6 < 30 /6

n<5

3 0

3 years ago

A canister is dropped from a helicopter 500m above the ground. Its parachute does not open, but the canister has been designed t

solmaris [256]

Answer:

The impact speed 98.995 m/s is less than 100 m/s and the canister will not burst.

Step-by-step explanation:

A function F is called an antiderivative of f on an interval I if $F'(x) = f(x)$ for all x in I.

Recall that if the object has position function $s=f(t)$ , then the velocity function is $v(t)=s'(t)$ . This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration function is $a(t)=v'(t)$ , so the velocity function is an antiderivative of the acceleration.

An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by $g$ . For motion close to the ground we may assume that $g$ is constant, its value being about $9.8 \:{\frac{m}{s^2}}$ .

We know that the acceleration due to gravity is given by

$a(t)=-9.8$

and the antiderivative is velocity

$v(t)=\int a(t)\,dt\\v(t)=\int -9.8\,dt\\v(t)=-9.8t +C$

We know that the canister was dropped, so the initial velocity at t = 0 is zero, this fact let us know the value of C.

$v(0)=9.8(0)+C\\C=0$

The antiderivative of velocity is the position

$s(t)=\int v(t) \, dt\\s(t)=\int -9.8t \, dt\\s(t)=-4.9t^2+C$

To find the value of the constant C, we know that the height was 500 m at t = 0, this means $s(0)=500$

$500=-4.9(0)^2+C\\C=500$

$s(t)=-4.9t^2+500$

Using the fact that at the time of impact the height s(t) is zero we can compute the total time of the fall:

$s(t)=-4.9t^2+500=0\\\\-4.9t^2=-500\\\\t^2=\frac{5000}{49}\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\t=\sqrt{\frac{5000}{49}},\:t=-\sqrt{\frac{5000}{49}}$

A negative time does not make sense, so we only take as a possible solution

$t=\sqrt{\frac{5000}{49}}=\frac{50\sqrt{2}}{7}\approx 10.102$

Now the final velocity is

$v(\frac{50\sqrt{2}}{7})=-9.8(\frac{50\sqrt{2}}{7})\approx -98.995$

The impact speed 98.995 m/s is less than 100 m/s and the canister will not burst.

5 0

4 years ago