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How do you solve -8x equals 14 equals -6x -30?
I'm an orange!
X plus 30 equals 3x-50 solve for x?
x+30 = 3x-50 x-3x = -50-30 -2x = -80 x = 40
How do you solve 13-3x equals 43?
13-3x = 43 -3x = 43-13 -3x = 30 x = -10
How do you Solve 6x equals 5over8?
you multiply it
Solve for x - x2-9x plus 14 equals 0?
x - x2 - 9x + 14 = 0 ; whence, x2 + 8x = 14 , x2 + 8x + 16 = 30 , and x + 4 = ±√30 . Therefore, x = ±√30 - 4 .
How do you solve -8x equals 14 equals -6x -30?
I'm an orange!
Solve 5x equals -25?
30
Solve 5 to the power of x equals -25?
30
How do you solve 0.75x equals 30?
0.75x = 30 Divide both sides by 0.75:- x = 40
Solve 45 equals 30 plus m?
45 = 30 + m Subtract 30 from both sides: 15 = m
How do you solve 7x plus 3y equals 30 -2x plus 3y equals 3 in an order pair?
(3,3)
How do you solve M 30 equals 30?
We don't know what operation you want. 1 x 30 = 30 0 + 30 = 30 60 - 30 = 30 900/30 = 30
How do you solve 30p equals 900?
30p = 900 divide each side by 30 p = 30 ------------ and 30 * 30 = 900 0r 30^2 = 900
Why are all even number digit palindromes divisible by 11?
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
If 30 plus 7 equals 37 how does that help you solve 37 plus 23?
It does not help you.
How do you solve 2v plus 11 equals -19 for v?
v is -15... -15*2 = -30... -30+11= -19.
Solve the system of equations 3x - 6y equals 30 y equals -6x plus 34?
x = 6 and y = -2
