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iren [92.7K]
2 years ago
14

Trains leave Bristol to Cardiff every 15 minutes to London every 21 minutes A train to Cardiff and a train to London both leave

Bristol at 11am. At what time will a train to Cardiff and a train to London next leave Bristol at the same time?
I don't know please help
​
Mathematics
1 answer:
artcher [175]2 years ago
7 0

Step-by-step explanation:

If we take lcm of 21 min and 15 min we get 105 min

therefore, 105 min after 11 a.m. would be 12:45 p.m.

hope this helps you

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What is the wavelength of a photon with an every of 3.38x 10^-19 J
Alexxandr [17]
From the attached graphic:
e = c*h/wavelength therefore,
wavelength = c*h / e
wavelength = 299,792,458 * 6.62606957e-34 / 3.38x10^-19
wavelength = 5.8771e-7 meters


4 0
3 years ago
Can anyone help with this?
PilotLPTM [1.2K]

Answer:

46

Step-by-step explanation:

The complete square starting with x^2+14x would be x^2+14x+49, since the square would be (x+7)^2. To make this square perfect, then, you would need to add 46 to make 49 with the 3. Hope this helps!

6 0
3 years ago
Read 2 more answers
FIRST TO ANSWER GETS BRALYIST<br> good luck :-)
Ratling [72]

Answer:

My answer is 20ft I'm not really sure but hope this helps

7 0
2 years ago
Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li
krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years

First, let us calculate the decay constant (k)

75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036

Next, let us calculate the half-life as follows:

\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2}  \approx 669\ years

Therefore the half-life is between 600 and 700 years

5 0
2 years ago
Find the distance between the points (6, 3) and (6,-5).
tiny-mole [99]

the answer is negative two over zero

6 0
2 years ago
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