Answer:
Option D - x = 3 3/5
Step-by-step explanation:
Lets make the left side into an improper fraction :
2 2/5 = 12/5
Now let's make both fraction into equivalent ones but with the same denominator :
2/3 = 10/15
12/5 = 36/15
Now we substitute the equivalent values back into the equation :
10/15 × x = 36/15
Now we multiply both sides by 15 to remove the denominators:
10 × x = 36
Now we divide both sides by 10 and simplify :
x = 36/10
x = 18/5
x = 3 3/5
So our answer will be Option D
Hope this helped and brainliest please
Hello,
the answer is D (said Al Kashi)
Imx->0 (asin2x + b log(cosx))/x4 = 1/2 [0/0 form] ,applying L'Hospital rule ,we get
= > limx->0 (2a*sinx*cosx - (b /cosx)*sinx)/ 4x3 = 1/2 => limx->0 (a*sin2x - b*tanx)/ 4x3 = 1/2 [0/0 form],
applying L'Hospital rule again ,we get,
= > limx->0 (2a*cos2x - b*sec2x) / 12x2 = 1/2
For above limit to exist,Numerator must be zero so that we get [0/0 form] & we can further proceed.
Hence 2a - b =0 => 2a = b ------(A)
limx->0 (b*cos2x - b*sec2x) / 12x2 = 1/2 [0/0 form], applying L'Hospital rule again ,we get,
= > limx->0 b*(-2sin2x - 2secx*secx.tanx) / 24x = 1/2 => limx->0 2b*[-sin2x - (1+tan2x)tanx] / 24x = 1/2
[0/0 form], applying L'Hospital rule again ,we get,
limx->0 2b*[-2cos2x - (sec2x+3tan2x*sec2x)] / 24 = 1/2 = > 2b[-2 -1] / 24 = 1/2 => -6b/24 = 1/2 => b = -2
from (A), we have , 2a = b => 2a = -2 => a = -1
Hence a =-1 & b = -2
It is a root of the polynomial if it causes the equation to equal 0.
Let's try each number.
Substituting -3 for x...
(-3)³ - 4(-3)² -11(-3) + 30
-27 - 36 + 33 + 30
0
Therefore, a. -3 is a root to our polynomial.
You could try the rest, but I won't waste time with it. None of the others are roots, and it's multiple choice which means usually just one answer.
