Study guides

☆☆

Q: How do you solve 10-2n equals 30?

Write your answer...

Submit

Still have questions?

Related questions

I'm an orange!

30

30

0.75x = 30 Divide both sides by 0.75:- x = 40

45 = 30 + m Subtract 30 from both sides: 15 = m

(3,3)

We don't know what operation you want. 1 x 30 = 30 0 + 30 = 30 60 - 30 = 30 900/30 = 30

30p = 900 divide each side by 30 p = 30 ------------ and 30 * 30 = 900 0r 30^2 = 900

It does not help you.

v is -15... -15*2 = -30... -30+11= -19.

x = 6 and y = -2

33

To solve this one, subtract 50 from 180 and you'll have your answer.

n=3,y=4 2(n+3y)=30

5/6=x/30 150=6x 25=x

2/6=x/30 60=6x 10=x

Using algebra to solve this problem (12 + .25x = 30), the answer (x) is 72.

11a+9 = 4a+30 11a-4a = 30-9 7a = 21 Solution: a = 3

x+30 = 3x-50 x-3x = -50-30 -2x = -80 x = 40

121 is the answer

First, by induction it is shown that 102n-1 ≡ -1 (mod 11) for all natural numbers n (that is, all odd powers of 10 are one less than a multiple of 11.) i) When n = 1, then we have 102(1)-1 = 10, which is certainly congruent to -1 (mod 11). ii) Assume 102n-1 ≡ -1 (mod 11). Then, 102(n+1)-1 = 102n+2-1 = 102×102n-1 102 ≡ 1 (mod 11) and 102n-1 ≡ -1 (mod 11), so their product, 102×102n-1, is congruent to 1×(-1) = -1 (mod 11). Now that it is established that 10, 1000, 100000, ... are congruent to -1 (mod 11), it is clear that 11, 1001, 100001, ... are divisible by 11. Therefore, all integer multiples of these numbers are also divisible by 11. A palindrome containing an even number of digits may always be written as a sum of multiples of such numbers. The general form of such a palindrome, where all the As are integers between 0 and 9 inclusive, is A0 100 + A1 101 + ... + An 10n + An 10n+1 + ... + A1 102n + A0 102n+1 which may be rewritten as A0 (100 + 102n+1) + A1 (101 + 102n) + ... + An (10n + 10n+1) and again as 100 A0 (1 + 102n+1) + 101 A1 (1 + 102n-1) + ... + 10n An (1 + 101) Each of the factors in parentheses is one more than an odd power of 10, and is hence divisible by 11. Therefore, each term, the product of one such factor with two integers, is divisible by 11. Finally, the sum of terms divisible by 11 is itself divisible by 11. QED

How do you solve 4y plus x equals 8

36

2x - 10 =30 2x -10 = 30 +10 +10 2x = 40 /2 /2 x = 20

x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}