Solve for m:3 m + 7/2 = 5/2 - 2 m
Put each term in 3 m + 7/2 over the common denominator 2: 3 m + 7/2 = (6 m)/2 + 7/2:(6 m)/2 + 7/2 = 5/2 - 2 m
(6 m)/2 + 7/2 = (6 m + 7)/2:(6 m + 7)/2 = 5/2 - 2 m
Put each term in 5/2 - 2 m over the common denominator 2: 5/2 - 2 m = 5/2 - (4 m)/2:(6 m + 7)/2 = 5/2 - (4 m)/2
5/2 - (4 m)/2 = (5 - 4 m)/2:(6 m + 7)/2 = (5 - 4 m)/2
Multiply both sides by 2:6 m + 7 = 5 - 4 m
Add 4 m to both sides:6 m + 4 m + 7 = (4 m - 4 m) + 5
4 m - 4 m = 0:6 m + 4 m + 7 = 5
6 m + 4 m = 10 m:10 m + 7 = 5
Subtract 7 from both sides:10 m + (7 - 7) = 5 - 7
7 - 7 = 0:10 m = 5 - 7
5 - 7 = -2:10 m = -2
Divide both sides of 10 m = -2 by 10:(10 m)/10 = (-2)/10
10/10 = 1:m = (-2)/10
The gcd of -2 and 10 is 2, so (-2)/10 = (2 (-1))/(2×5) = 2/2×(-1)/5 = (-1)/5:Answer: m = (-1)/5
The rule of law of the square root of constants is taking the nearest divisible number which has a square root of an integer. In this case, 50 is divisible by 25. Hence sqrt of 50 is 5 sqrt 2. The sqrt of a variable applies the multiplication rule. x^6 is raised to 0.5 equal to x^3. Hence the answer is 5 x3 sqrt 2.
Hello :
(-6a)-(-3a) =(-3a)-(0) = (0)-(3a) = (3a)-(6a) = -3a (arthmetic<span> sequence, the common - diff is : d= -3a the first term is : U1 = 6a
</span><span>a general rule for the n th term is : Un =U1+(n-1)d
Un = 6a +(n-1)(-3a) =
Un = 6a -3an+3a
</span>Un = -3an +9a
Answer:
The correct answer is 7.
Step-by-step explanation:
Let there be x number of candies in each bag filled by Ken and Mali.
Both Ken and Mali started with equal number of candies and put same number of candies in each bag.
Ken filled nine bags and had three candies left. Therefore total number of candies Ken had = 9x + 3.
Mali filled eight bags and had ten candies left. Therefore total number of candies Mali had = 8x + 10.
As both had equal number of candies in the beginning,
9x + 3 = 8x + 10
⇒ x = 10 - 3 = 7
Thus each bag contained 7 candies.
Answer:
doesn't it give you the angles for each letter? If that is the case then it would be 33
Step-by-step explanation:
May I have brainliest please? :)
