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Jobisdone [24]
3 years ago
11

Susan and teo play soccer. Susan's team play every five days and twos team plays every six days when will the teams play on the

same day
Mathematics
1 answer:
Serjik [45]3 years ago
7 0
You're looking for the least common multiple of the numbers 5 & 6

that is . . . you're looking for the first time that a multiple of the value coincides with a multiple of the second value

multiples of 5
5
10
15
20
25
30
35
40
 . . .

multiples of 6
6
12
18
24
30
36
42
 . . .

as you can see, the first common multiple of both 5 & 6 is 30, so the answer is <u><em>on the 30th day</em></u>
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Need help with this geometric question ASAP, Thank you
Law Incorporation [45]
The error would be assuming the altitude is a bisector and divides the sides evenly.
7 0
2 years ago
(t+9)20=140 what is t
GarryVolchara [31]

Answer:

t=-2

Step-by-step explanation:

(t+9)20=140

Distribute the 20

20t +180 = 140

Subtract 180 from each side

20t +180-180 = 140-180

20t = -40

Divide by 20

20t/20 = -40/20

t=-2

8 0
3 years ago
Read 2 more answers
A second important result is that electrons will fill the lowest energy states available. This would seem to indicate that every
Assoli18 [71]

Answer:

  • 8

Explanation:

1)<u> Principal quantum number, n = 2</u>

  • n is the principal quantum number and indicates the main energy level.

<u>2) Second quantum number, ℓ</u>

  • The second quantum number, ℓ,  is named, Azimuthal quantum number.

The possible values of ℓ are from 0 to n - 1.

Hence, since n = 2, there are two possible values for ℓ: 0, and 1.

This gives you two shapes for the orbitals: 0 corresponds to "s" orbitals, and 1 corresponds to "p" orbitals.

<u>3) Third quantum number, mℓ</u>

  •    The third quantum number, mℓ, is named magnetic quantum number.

The possible values for mℓ are from - ℓ to + ℓ.

Hence, the poosible values for mℓ when n = 2 are:

  • for ℓ = 0: mℓ = 0
  • for ℓ = 1, mℓ = -1, 0, or +1.

<u>4) Fourth quantum number, ms.</u>

  • This is the spin number and it can be either +1/2  or -1/2.

Therfore the full set of possible states (different quantum number for a given atom) for n = 2 is:

  • (2, 0, 0 +1/2)
  • (2, 0, 0, -1/2)
  • (2, 1, - 1, + 1/2)
  • (2, 1, -1, -1/2)
  • (2, 1, 0, +1/2)
  • (2, 1, 0, -1/2)
  • (2, 1, 1, +1/2)
  • (2, 1, 1, -1/2)

That is a total of <u>8 different possible states</u>, which is the answer for the question.

8 0
3 years ago
At a frisbee throwing competition one contestant threw a frisbee 113.47 meters
jekas [21]
I don’t understand the question
5 0
3 years ago
Find all zeros x^3+2x^2+4x+21
dlinn [17]

Answer:

see explanation

Step-by-step explanation:

note when x = - 3

(- 3)³ + 2(- 3)² + 4(- 3) + 21 = - 27 + 18 - 12 + 21 = 0

hence x = - 3 is a zero and (x + 3) is a factor and dividing gives

\frac{x^3+2x^2+4x+21}{x+3} = (x + 3)(x² - x + 7)

For zeros equate to zero

(x + 3)(x² - x + 7) = 0

equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x² - x + 7 = 0 ← solve using quadratic formula

x = (1 ± \sqrt{1-28} ) / 2 = (1 ± 3i\sqrt{3} ) / 2

x = \frac{1}{2} ± \frac{3i\sqrt{3} }{2}

zeros are x = - 3, x = \frac{1}{2} ±\frac{3i\sqrt{3} }{2}


5 0
3 years ago
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