a class of 25 students shares a class set of 100 markers. on a day with 5 students absent, which statement is true
2 answers:
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9 longer sticks were used.
This is how i get it.
1. change m to cm, 1m =100cm
2. add both together, divide by 16 =8.125. Which means 1 of the stick should more than 8, and 1 of the other stick should be less than 8
3. substitute the equation
4.(55 x 7)+(75 x 9) =1060
This is how i get it.
1. change m to cm, 1m =100cm
2. add both together, divide by 16 =8.125. Which means 1 of the stick should more than 8, and 1 of the other stick should be less than 8
3. substitute the equation
4.(55 x 7)+(75 x 9) =1060
Answer:
50
Step-by-step explanation:
and
can be expressed in complex form, with
= i
=
=
×
= 8i
=
=
×
= 4i
the factors can then be expressed as
(6 + 8i)(3 - 4i) ← expand using FOIL
= 18 - 24i + 24i - 32i² [ i² = ( )² = - 1 ]
= 18 - 24i + 24i + 32 ← collect like terms
= 18 + 32 + 0
= 50
Let a be the first term in the sequence, and d the common difference between consecutive terms. If aₙ denotes the n-th term in the sequence, then
a₁ = a
a₂ = a₁ + d = a + d
a₃ = a₂ + d = a + 2d
a₄ = a₃ + d = a + 3d
and so on, up to the n-th term
aₙ = a + (n - 1) d
The sum of the first 10 terms is 100, and so
where we use the well-known sum formulas,
The sum of the next 10 terms is 300, so
Solve for a and d. Eliminating a gives
(10a + 145d) - (10a + 45d) = 300 - 100
100d = 200
d = 2
and solving for a gives
10a + 145×2 = 300
10a = 10
a = 1
So, the given sequence is simply the sequence of positive odd integers,
{1, 3, 5, 7, 9, …}
given recursively by the relation
and explicitly by
for n ≥ 1.