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How do you solve for m plus n equals 1000 and 0.05m plus 0.06n equals 57?
Solve be substitution... m+n=1000 0.05m+0.06n=57 Take one of the equation and find "m" m=1000-n substitute into the other equation 0.05(1000-n)+0.06n=57 50-0.05n+0.06n=57 50-0.01n=57 -0.01n=-7 n=700 Now you have found "n" now find "m" m=1000-n m=1000-700 m=300 m=300 and n=700
How many n in a kn?
There are 1000 of them.
How much 15 kN in to N?
15 kN = 15 x 1000 N = 15000 N
How many positive integers n from 1 to 1000 inclusive are there such that n is a multiple of 3 and the digit sum of n is also a multiple of 3?
All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.
If you add up all the numbers from 1 to 1000 what do you get?
Sum of first n numbers = n/2(n +1) = 500 x 1001 = 500500
If n times 1000 is equal to 0.6 what is n?
1000n = 0.6 Therefore, 0.6/1000 = n Therefore, n = 0.0006
When was Hard 'n' Heavy created?
Hard 'n' Heavy was created on 1981-05-25.
When was The Hot n' Heavy created?
The Hot n' Heavy was created on 2009-06-02.
How heavy is the spaceship?
about 1000 tons
Find the programming code for calculating the sum of the squares of the first 1000 numbers in HASKELL?
To get a list of the squares of the first 1000 numbers we can do:> [n^2 | n sum [n^2 | n
What is One kilogram is times what as heavy as one gram?
1000 times as heavy.
How heavy is 1000 pounds in kilograms?
453.59237
N 1000 equals 0.6?
N = .0006
How heavy is one metric ton?
1 metric tonne is 1000 kilograms. Is that heavy?
A 90 kg weightlifter exerts a 1000 N force upwards against a 50 kg barbell How large is the NET vertical force acting on the barbell?
The weightlifter exerts a force of 1000 N upwards, and gravity exerts a force downward on the weightlifter and the barbell. The net force acting on the barbell is the difference between the force exerted by the weightlifter and the force of gravity acting on both the weightlifter and the barbell, which is 1000 N - (90 kg + 50 kg) * 9.8 m/s^2.
What is 1000 plus 999 plus 998 etc to 1?
There is an equation to find the sum of a list of numbers . It is called the Arithmetic Progression. The equation is S(n) = (n/2)[a + (n-1)d] 4 Where n = number of terms (1000) a = first term (1000) d = difference between terms (1) Hence S(1000) = (1000/2)[1000 + ( 1000 - 1)(1)] S(1000) = 500[ 1000 + 999] S(1000) = 500[1999] S(1000) = 999500 The answer!!!!!
When did N-Sub happen?
N-Sub happened in 1000.
