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To solve the equation 10n-3 = 37, the first thing we do is add 3 to both sides. This gives us 10n = 40. We then divide both sides by 10, yielding the result of n=4. To prove the equation, we plug in the number 4: 10 (4)-3 = 37, which gives us 40-3=37, which is 37=37, so 4 is the correct answer.
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What is n divided by 3 7?
It is n/37
If 15 is proportional to 37 then what is proportional to 50?
15/37 = n/50 ie n = 15 x 50/37 = 20&10/37 or 20.270 all recurring Alternatively 37/15 = 50/n which gives the same solution.
What are four consecutive integers with the sum of -142?
Let the first of the four consecutive integers be represented by n, then the other three can be (n + 1), (n + 2) and (n + 3) Then we have, n + (n + 1) + (n + 2) + (n + 3) = -142 which simplifies to, 4n + 6 = -142 so, 4n = -142 - 6 = -148........n = -148/4 = -37 The four integers are thus, -37, -36 (-37 + 1), -35 (-37 + 2) and -34 (-37 + 3) and we have, -37 +(-36) + (-35) + (-34) = -142
What is n plus five and three tenths equals eight and one tenths?
n+5+(3/10)=8+(1/10) n+5-5+(3/10)=8-5+(1/10) n+(3/10)=3+(1/10) n+(3/10)=2+(11/10) n+(3/10)-(3/10)=2+(11/10)-(3/10) n=2+(8/10) n is equal to 2 and 8 tenths
Solve by the substitution method 5m plus n equals 37 m-2n equals 14?
5m+n = 37 m-2n = 14 => m = 2n+14 Substitute the value of m into the top equation: 5(2n+14)+n = 37 10n+70+n = 37 10n+n = 37-70 11n = -33 Divide both sides by 11 to find the value of n: n = -3 Substitute the value of n into the original equations to find the value of m: Therefore: n = -3 and m = 8
What is 3 consecutive number when the product of the smaller two number is 37 less than the square of the largest number?
Let the 3 consecutive number be represented by n, (n + 1) and (n + 2) Then, n(n + 1) + 37 = (n + 2)2 n2 + n + 37 = n2 + 4n + 4 3n = 33 n = 11 The three numbers are 11, 12 and 13.
4 the sum of a number and - 37?
-33
How do you simplify n minus 10 plus 9n minus 3?
n - 10 + 9n - 3 n + 9n - 10 - 3 10n - 13
How many combinations of 10 numbers can you get from 40?
The answer is 40!/[10!*(40-10)!] where n! represents 1*2*3*...*n. The number of combinations = 40*39*38*37*36*35*34*33*32*31/(10*9*8*7*6*5*4*3*2*1) = 847,660,528
How do you write 1 x 3 7 x 3 as a product of two factors?
If your number N is equal to N = w*x*y*z, then you could also write it as N = a*y*z, where a = w*x.
What is the next two patterns of 2 5 10 17?
Each successive number after 2, is formed by adding 3, then 5, then 7, and so on. 2 + 3 =, 5 5 + 5 = 10 10 + 7 = 17 So the next two numbers are : 17 + 9 = 26 26 + 11 = 37 The formula for the series is n2 + 1 So when n = 1, n2 + 1 = 2 When n = 5, n2 + 1 = 26 When n = 6, n2 + 1 = 37..........and so on
How do you work out the third term of 10 subtract n squared?
To solve the third term of 10-n², substitute the number 3 for n. 10-n² =10-3² =10-9 =1 The third term of 10-n² is 1.
