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suter [353]
3 years ago
15

Jonathan is in a class of 15 boys and 35 girls.40%of the students in the class take the bus to school how many students do not t

ake the bus to school ?
Mathematics
2 answers:
sertanlavr [38]3 years ago
7 0
40% of 40
10% of 40 is 4
4 x 4=16
40-16 =24
Answer is 24
Svetllana [295]3 years ago
6 0
15 + 35 to find the total of students in the class which is 50.
100 / 40 = 2.5, so divide 50 by 2.5 and you get 20

Answer is 20
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5000

Step-by-step explanation:

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What are the zeros for both functions
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Answer:

the zero of a function is the value of x which makes the final value zero

First Equation:

So let 25 - 2x equal to 0

<em>25 - 2x = 0</em>

x = 12.5

Second Equation:

Let 2x² - 11x - 6 equal to zero

<em>2x² - 11x - 6 = 0 </em>

<em>2x² - 12x + x - 6 = 0 </em><em>(Splitting the middle term)</em>

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<em>(2x + 1) (x - 6) = 0</em>

So we can transpose either one of the brackets below the zero

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5 0
3 years ago
In ABC, C is a right angle, what is the measure
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In quadratic drag problem, the deceleration is proportional to the square of velocity
Mars2501 [29]
Part A

Given that a= \frac{dv}{dt} =-kv^2

Then, 

\int dv= -kv^2\int dt \\  \\ \Rightarrow v(t)=-kv^2t+c

For v(0)=v_0, then

v(0)=-kv^2(0)+c=v_0 \\  \\ \Rightarrow c=v_0

Thus, v(t)=-kv(t)^2t+v_0

For v(t)= \frac{1}{2} v_0, we have

\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\  \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\  \\ \Rightarrow kv_0t=2 \\  \\ \Rightarrow t= \frac{2}{kv_0}


Part B

Recall that from part A, 

v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\  \\ \Rightarrow dx=-kv^2tdt+v_0dt \\  \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\  \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a

Now, at initial position, t = 0 and v=v_0, thus we have

x=a

and when the velocity drops to half its value, v= \frac{1}{2} v_0 and t= \frac{2}{kv_0}

Thus,

x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\  \\ =- \frac{1}{2k} + \frac{2}{k} +a

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

- \frac{1}{2k} + \frac{2}{k} +a-a \\  \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}
7 0
3 years ago
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