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lbvjy [14]
4 years ago
8

Suppose it took 19 riders a total of 11 days 21 hours to ride from st Jospeh to Sacramento. If they all rode the same number of

hours, how many hours did each rider ride
Mathematics
1 answer:
ale4655 [162]4 years ago
6 0
285 hours. Did the riders travel together? If so the answer is 285 for each rider, because they were all together. If they like made shifts or something it would have been 15 hours. 
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The difference of two polynomials must always result in a _____.(1 point)
Mrrafil [7]

Answer:

polynomial...

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3 years ago
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At the end of April, Mandy told Bill that she has read 16 books this year and reads 2 books each month. Bill wants to catch up t
klemol [59]
Mandy Increases her books by 2 per month.
Bill increases his books by 4 per month.

Month     Mandy    Bill
May          18          4
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July          22        12
August     24        16
Sept        26        20
Oct          28        24
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At the end of December they will have read the same amount of books.
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3 years ago
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Answer:720

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3 0
3 years ago
The half-life of a certain radioactive substance is 45 days. There are 6.2 grams present initially. On what day
kvasek [131]

Answer:

There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days

Step-by-step explanation:

The given parameters are;

The half life of the radioactive substance = 45 days

The mass of the substance initially present = 6.2 grams

The expression for evaluating the half life is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}

Where;

N(t) = The amount of the substance left after a given time period = 1 gram

N₀ = The initial amount of the radioactive substance = 6.2 grams

t_{1/2} = The half life of the radioactive substance = 45 days

Substituting the values gives;

1 = 6.2 \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}

\dfrac{1}{6.2}  =  \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}

ln\left (\dfrac{1}{6.2} \right )  =  {\dfrac{t}{45} \times ln \left (\dfrac{1}{2} \right )

t = 45 \times \dfrac{ln\left (\dfrac{1}{6.2} \right ) }{ln \left (\dfrac{1}{2} \right )} \approx 118.45 \ days

The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days

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3 0
3 years ago
A park has a 333 meter (\text{m})(m)left parenthesis, start text, m, end text, right parenthesis tall tether ball pole and a 6.8
dybincka [34]

Answer:option B & D are correct

Step-by-step explanation:

Given options are  

                                             A              B          C             D           E        

Tether ball pole shadow      1.35         1.8        3.75        0.6       2

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3 meter tall tether ball pole and a 6.8 tall flagpole

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C    8.35/3.75  = 835/375  =  167/75   ≠ 34/15( 170/75)

D  1.36/0.6  = 136/60   =  34/15  = 34/15   ( Correct option)

E  4.8/2  = 48/20   =  12/5   = 36/15    ≠ 34/15

option B & D are correct

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3 years ago
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