F is continuous every-where. But it is not differentiable at two points. Note <span>it is not differentiable at </span><span>x = 0</span><span> and </span><span>x = 1! I mean just consider </span>f(x)= ∣x∣+∣x − 1∣
If you need me to expand let me explain it a bit more.
If you define a function f(x) so that f(x) = |x| for x<0. f(x) = sinx for 0 ≤ x < 22/7, We have f(x) = |x - 22/7| for x ≥ 22/7 then we can say this function is continuous however it will still have two points that are not differentiable. This would be for |x|
4x = -60 - 19y
-7x = -48 - 19y
Subtract the bottom equation from the top:
4x + 7x = -60 + 48 - 19y + 19y
Simplify:
11x = -12 -0y
11x = -12
Divide both sides by 11:
11x/11 = -12/11
Simplify:
x = -12/11
Then plug in x to solve for y:
4(-12/11) = -60 - 19y
Simplify:
-48/11 = -60 - 19y
Add 60 to both sides (keep in mind 60 = 660/11):
-48/11 + 660/11 = -60 + 60 - 19y
Simplify:
612/11 = 0 - 19y
612/11 = -19y
Divide both sides by -19 (keep in mind that dividing by -19 is the same as multiplying by -1/19):
612/11 • -1/19 = -19y/-19
Simplify:
-612/209 = y
y = -612/209
So, the answer is: (-12/11, -612/209)
Answer:
according to BODMAS rule
Step-by-step explanation:
a=17
b=4
c=12