Answer: 334
Step-by-step explanation:
6 consecutive numbers can be written as:
n, n+1, n+2, n + 3, n + 4, n + 5,
The addition of those 6 numbers is:
n + n+1 + n+2 + n + 3 + n + 4 + n + 5
6n + 1 + 2 + 3 + 4 + 5 = 6n + 15
Let's find the maximum n possible:
6n + 15 = 2020
6n = 2020 - 15 = 2005
n = 2005/6 = 334.16
The fact that n is a rational number means that 2020 is can not be constructed by adding six consecutive numbers, but we know that with n = 334 we can find a number that is smaller than 2020, and with n = 335 we can found a number bigger than 2020.
So with n = 334 we can find one smaller.
6*334 + 15 = 2019
and we can do this for all the values of n between 1 and 334, this means that we have 334 numbers less than 2020 that can be written as a sum of six consecutive positive numbers.
Answer:
The displacement is: - 80 i
Step-by-step explanation:
In order to answer the question, you have to apply the displacement formula, which is:
ΔX= Xf-Xi
Where ΔX is the displacement, Xf is the final position and Xi is the initial position. The formula represents the change in the position of an object from the origin.
Let the origin be in 0 km,with the x-axis positive to te right.
Notice that the displacement is a vector. The positions are:
Xf=163 km (i)
Xi= 243 km (i)
Because John starts in 243 km and finishes in 163 km.
Where i is the unit vector in the x direction.
Therefore:
ΔX= 163 i - 243 i
Solving:
ΔX= -80 i
Notice that for calculating the displacement you just need the initial and final position. It doesn't depend on the distance traveled.
They can show change on how much time passed.
Answer:
see below
Step-by-step explanation:
3x^2 + 17x +10 =0
(3x+ ) ( x+ ) =0
We need to factor 10 = 2*5 or 1*10
They are positive since we have all positive terms
3*5 = 15+2 = 17
The 5 needs to multiply the 3 so it goes on the outside
(3x+ ) ( x+ 5 ) =0
That leaves the 2 on the inside
(3x+ 2 ) ( x+ 5 ) =0
Using the zero product property
3x+2 =0 x+5=0
3x = -2 x=-5
x = -2/3 x = -5
Answer: n = 130
Step-by-step explanation:
The sequence is an Arithmetic progression ( AP )
The last term of the sequence is (L) is 394 with the formula
L = a + ( n- 1 )d . From the sequence, a = 7, d ( common difference ) = 3 and L ( last term ) = 394, and n = ?
Put those values in the formula above and solve for n.
a + ( n - 1 )d = L
7 + ( n - 1 ) x 3 = 394
7 + 3n - 3 = 394
4 + 3n = 394
3n = 394 - 4
3n = 390
n = 130
