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amm1812
4 years ago
9

Find the missing side of each triangle. leave your answers in simplest radical form

Mathematics
1 answer:
dedylja [7]4 years ago
7 0
X^2 = 4^2 - (2√3)^2
x^2 = 16 - 12
x^2 = 4
x = 2

answer
x = 2 cm
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Help me plz I need it I'm very desperate help
Klio2033 [76]

Answer:

1/45

Step-by-step explanation:

There are 10 marbles

P ( yellow) = yellow marbles / total = 2/10 = 1/5

We do not replace the marble

There are 9 marbles, 1 of which is yellow

P ( yellow) = yellow marbles / total = 1/9

P ( yellow, no replacement, yellow) = 1/5 * 1/9 = 1/45

8 0
4 years ago
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Solve 3y^{2} - (y + 2) (y - 2)
hram777 [196]

Answer:

3y^2 - (y + 2) (y - 2) = 0

<=> 3y^2 - (y^2 - 4) = 0

<=> 2y^2 + 4 =0

<=>y^2 + 2 = 0

=> Because y^2 is always equal or larger than 0, there is no real solution.

Hope this helps!

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4 years ago
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What is the value of x? <br><br><br> º
Sonbull [250]
The answer would be 65

8 0
3 years ago
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
What is the LCM of 33y^3, 33y, 22x^2y
lidiya [134]
LCM - means least common multiple
3 0
3 years ago
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