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3 years ago

How to find square root of a big number

Mathematics

1 answer:

snow_tiger [21]3 years ago

7 0

U can do that number^(1/2)

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For the following true conditional statement, write the converse. If the converse is also true, combine the statements as a bico

Licemer1 [7]

Original Statement: If x = 3, then x^2 = 9 (true)

Converse: If x^2 = 9, then x = 3 (false; this is half the story and x = -3 must be included)

Since the converse isn't completely true, we can't combine the two statements to get a biconditional.

5 0

3 years ago

Three more than the product of two and a number x

AfilCa [17]

The equation the statement shows is 2x+3

5 0

3 years ago

Read 2 more answers

The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

Please someone help me with number 21 part A

Vinvika [58]

Because since 17 can only be multiplied by it's self and one that's what the box is showing

6 0

3 years ago

Anyone know how to do this?¿?¿

vichka [17]

0.65 x 40 = 26
So the answer is 26 pounds.

7 0

3 years ago

Read 2 more answers