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GenaCL600 [577]
3 years ago
15

The radius of a nitrogen atom is 5.8*10^-11 meters, and the radius of a beryllium atom is 1.12*10^-10 meters. Which atom has a l

arger radius, and by how many times is it larger?
Mathematics
1 answer:
Burka [1]3 years ago
5 0
Nitrogen Radius = 5.8 x 10⁻¹¹ m

Beryllium Radius = 1.12 x 10⁻¹⁰ m

Let's find the quotient of N/Be :

(5.8x10⁻¹¹)/(1.12x10⁻¹⁰). But 10⁻¹¹/10⁻¹⁰ = 10⁽⁻¹¹⁺¹⁰⁾ = 10⁻¹ = 1/10 = 0.1

→ (5.8/1.12).(0.1) = 0.58/1.12 = 0.518.

Conclusion: the radius of Be is almost double than the radius of N
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What is the focus of the parabola? y=−1/4x2−x+3
alex41 [277]

Answer:  Focus = (-2, 3)

<u>Step-by-step explanation:</u>

y=-\dfrac{1}{4}x^2-x+3\\\\\rightarrow a=-\dfrac{1}{4},\ b=-1

First let's find the vertex. We do that by finding the Axis-Of-Symmetry:

AOS: x=\dfrac{-b}{2a}\quad =\dfrac{-(-1)}{2(\frac{-1}{4})}=\dfrac{1}{-\frac{1}{2}}=-2

Then finding the maximum by inputting x = -2 into the given equation:

y=-\dfrac{1}{4}(-2)^2-(-2)+3\\\\y=-1+2+3\\\\y=4

The vertex is: (-2, 4)

Now let's find p, which is the distance from the vertex to the focus:

a=\dfrac{1}{4p}\\\\\\-\dfrac{1}{4}=\dfrac{1}{4p}\\\\\\p=-1

The vertex is (-2, 4) and p = -1

The focus is (-2, 4 + p) = (-2, 4 - 1) = (-2, 3)

3 0
3 years ago
A company offers a raffle whose grand prize is a $35,000 new car. Additional prizes are a $1,100 television and $600 computer. T
anastassius [24]

Answer:

-$ 10.13

Step-by-step explanation:

Given,

The cost price of each ticket = $ 22,

So, the value of each ticket other than price ticket = - $ 22, ( negative sign shows loss,i.e. if we don't get the price we will have the loss of $ 22 )

Now, there are 3 tickets in which first is of $35,000 price, second is of $1,100 and third is of $ 600,

So, the value of first ticket = 35000 - 22 = $ 34978,

Value of second ticket = 1100 - 22 = $ 1078,

Value of third ticket = 600 - 22 = $ 578,

Also,

\text{Probability}=\frac{\text{Favourable outcome}}{\text{Total outcome}}

Hence, by the above information we can make a table for the given situation,

Number           2997               1                    1                     1        

Price               -$ 22               $ 34978        $ 1078           $ 578

Probability       2997/3000      1/3000         1/3000           1/3000

Therefore, the expected value of a ticket

=-22\times \frac{2997}{3000}+34978\times \frac{1}{3000}+578\times \frac{1}{3000}

=-$10.126

≈ - $ 10.13

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3 years ago
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