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OLga [1]
3 years ago
13

Can anyone help? I’ve posted this like 4 times now plz if u can. A bicyclist traveled with a speed of 12 km/h from a camping gro

und to a station, located 60 km away. Write a formula expressing the dependence of the variable s on t, where s is the distance between the bicyclist and the station in km, t - the duration (time) of his travel in hours. Using the formula, find:t for s=30I’ve already got the time, I just need the formula
Mathematics
1 answer:
Tresset [83]3 years ago
7 0
Here is it


s = 60 - 12t

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Find how many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6 and 7 (with repetitions), if:
Goshia [24]

Answer:

case 1 = 2592

case 2 =  729

case 1 + case 2 =  2916

(this is not a direct adition, because case 1 and case 2 have some shared elements)

Step-by-step explanation:

Case 1)

6 digits numbers that can be divided by 25.

For the first four positions, we can use any of the 6 given numbers.

For the last two positions, we have that the only numbers that can be divided by 25 are numbers that end in 25, 50, 75 or 100.

The only two that we can create with the numbers given are 25 and 75.

So for the fifth position we have 2 options, 2 or 7,

and for the last position we have only one option, 5.

Then the total number of combinations is:

C = 6*6*6*6*2*1 = 2592

case 2)

The even numbers are 2,4 and 6

the odd numbers are 3, 5 and 7.

For the even positions we can only use odd numbers, we have 3 even positions and 3 odd numbers, so the combinations are:

3*3*3

For the odd positions we can only use even numbers, we have 3 even numbers, so the number of combinations is:

3*3*3

we can put those two togheter and get that the total number of combinations is:

C = 3*3*3*3*3*3 = 3^6 = 729

If we want to calculate the combinations togheter, we need to discard the cases where we use 2 in the fourth position and 5 in the sixt position (because those numbers are already counted in case 1) so we have 2 numbers for the fifth position and 2 numbers for the sixt position

Then the number of combinations is

C = 3*3*3*3*2*2 = 324

Case 1 + case 2 = 324 + 2592 = 2916

4 0
2 years ago
Find [5(cos 330 degrees + I sin 330 degrees)]^3
earnstyle [38]
Given a complex number in the form:
z= \rho [\cos \theta + i \sin \theta]
The nth-power of this number, z^n, can be calculated as follows:

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z^n = \rho^n [\cos n\theta + i \sin n\theta ]
In our case, n=3, so z^3 is equal to
z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ] (1)
And since 
3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}
and both sine and cosine are periodic in 2 \pi,  (1) becomes
z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]

6 0
3 years ago
Ten times as much as 3000
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Multiply 3000 by 10

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3 0
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monitta
Area of a circle= πr²

Data:
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Answer:

not sure i hope you find someone to help u tho soon

Step-by-step explanation:

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3 years ago
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