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zheka24 [161]
3 years ago
15

There are 8 students on a minibus. 5 students are boys. What is the fraction of students?

Mathematics
2 answers:
JulijaS [17]3 years ago
6 0
5/8, because there are 8 people and 5 boys.
pav-90 [236]3 years ago
4 0
The fraction of boys to total students would be 5/8

The fraction of boys to girls would be 5/3
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The work of a student to solve the equation 2(5y – 2) = 12 + 6y is shown below: Step 1: 2(5y – 2) = 12 + 6y Step 2: 7y – 4 = 12
kirza4 [7]
2(5y - 2) = 12 + 6y
7y - 4 = 12 + 6y ⇒ wrong, it has to be: 10y - 4 = 12 + 6y

so the correct answer is B 
7 0
3 years ago
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An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 

substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
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4 0
3 years ago
A series of three? separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers. Each
Nadya [2.5K]

Answer:

312.5\pi \text{ km}^3\approx 981.75\text{ km}^3

Step-by-step explanation:

We have been given that a series of 3 separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers.

Each of the three tunnels is shaped like a half-cylinder with a radius of 5 meters.

Since we know that volume of a semicircular or a half cylinder is half the volume of a circular cylinder.

\text{Volume of a semicircular cylinder}=\frac{\pi r^2h}{2}, where,

r = Radius of cylinder,

h = height of the cylinder.

Upon substituting our given values in volume formula we will get,

\text{Volume of a semicircular cylinder}=\frac{\pi (5\text{ km})^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*25\text{ km}^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*625\text{ km}^3}{2}

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=981.74770\text{ km}^3

Therefore, the volume of earth removed to build the three tunnels is 312.5\pi \text{ km}^3\approx 981.75\text{ km}^3.


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your answer is A :)
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