Answer:
The Consecutive number are 46 , 47 and 48 having sum as 141
The smallest number is 46
Step-by-step explanation:
• A consecutive numbers means number are in a sequence and are one after the another !
e.g. n , (n+1). Or 1 , 2 are consecutive numbers
• A consecutive even numbers differ by 2 because all even numbers are divisible by 2 !
• If the number is divisible by x we must had taken number as n , ( n + x ) ,
( n + 2x )
• Let the numbers be ( n - 1 ) , n and
( n + 1 )
Given ,the sum of numbers is 141
n - 1 + n + n + 1 = 141
3n = 141
n = 141/3
n = 47
So, the numbers are
n- 1 = 47 - 1 = 46
n = 47
n + 1 = 48
And , the smallest number is 46
• Verification :-
46 + 48 + 47 = 141
Which satisfies the given condition !!
Hence , verified !!
2(z - 5) + (z - 8) =
= 2z - 10 + z - 8 =
= 2z + z - 10 - 8 = <u>3</u><u>z</u><u> </u><u>-</u><u> </u><u>1</u><u>8</u> ← the end
For this case we have that by definition, the GCF of two numbers is the biggest common factor that divides both numbers without leaving residue. We find the factors:
26: 1,2,13,26
78: 1,2,3,6,13,26
Thus, the GCF of both numbers is 26
Answer:
26
Option C
So,first step is to write ![(fog)(-4)) =f[g(-4)] \\](https://tex.z-dn.net/?f=%20%28fog%29%28-4%29%29%20%3Df%5Bg%28-4%29%5D%20%5C%5C%20)
Now we start from inner paranthesis
,we need to first find value of
![g(-4) =\frac{\sqrt{12-(-4)}}{3}\\ =\frac{\sqrt{12+4}}{3}\\ =\frac{\sqrt{16}}{3}\\ =\frac{4}{3}\\ (fog)(-4)) =f[g(-4)] =f(\frac{4}{3}) =9(4/3)^{2} =9*(16/9) =144/9 =16](https://tex.z-dn.net/?f=%20g%28-4%29%20%3D%5Cfrac%7B%5Csqrt%7B12-%28-4%29%7D%7D%7B3%7D%5C%5C%20%3D%5Cfrac%7B%5Csqrt%7B12%2B4%7D%7D%7B3%7D%5C%5C%0A%3D%5Cfrac%7B%5Csqrt%7B16%7D%7D%7B3%7D%5C%5C%0A%3D%5Cfrac%7B4%7D%7B3%7D%5C%5C%0A%28fog%29%28-4%29%29%20%3Df%5Bg%28-4%29%5D%20%3Df%28%5Cfrac%7B4%7D%7B3%7D%29%20%3D9%284%2F3%29%5E%7B2%7D%20%3D9%2A%2816%2F9%29%20%3D144%2F9%20%3D16%20)
