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Blababa [14]
3 years ago
8

Ok what is 84% of 311?

Mathematics
2 answers:
kobusy [5.1K]3 years ago
8 0
It would be: 311*84 / 100 = 311 * 0.84 = 261.24
So, your answer is 261.24
erastova [34]3 years ago
8 0
84\100 x 311
84 x 311=26124\100
answer=261.24
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Fifty balls numbered 1 through 50 are mixed up in a barrel. How many balls must you draw from the barrel (without looking) to be
ipn [44]
If you draw only 25 balls, you could draw the 25 odd-numbered balls.  However, there will then be no odd-numbered balls left, so when you draw two more you will be guaranteed to get two even-numbered balls.  Thus, the minimum is 25+2=27 balls.
8 0
3 years ago
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
How do you solve each system of equations?
stiv31 [10]

Answer:

x=1,y=3,z=1

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4x-4y+4z=-4

4x=-4+4y-4z

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Substitute z=1 into y=\frac{6z+9}{5}:

y=3

Plug in y and z values into x=-1+y-z:

x=1

4 0
3 years ago
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prisoha [69]

Answer:   888+88+8+8+8 = 1,000

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6 0
3 years ago
Can someone pls answer this question what is 12 +z=15
ANTONII [103]

12 + z = 15


12 - 12 + z = 15 - 12


0 + z = 3


z = 3


3 0
3 years ago
Read 2 more answers
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