The line integral is the path of the function along a line having a continuous value.
The integral solved gives the value 111.
The given question is
where C is the line from (1, 0, 0) to (4, 1, 3)
Here
x→ 1⇒ 4
y→ 0⇒1
z→ 0⇒ 3
Let t be defined be the range 0≤ t ≤ 1
Then x= 4t+1 : dx= 4dt
y= t ": dy= dt
z= 3t : dz= 3dt
Putting the values
= [36(1)²- 36(0)²]+ 3 [16(1)² +1+8(1)] - 3 [16(0)² +1+8(0)] + [3(1)²- 3(0)² ]
= 36 +75-3+3
= 111
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Any variable raised to the zero power is 1
y = x^0 is 1
Even any number raised to the power of 0 is 1.
5555^0 =1
Well not every number. 0^0 is a bit of an exception. 0^0 is undefined and you for that reason you should be careful how you treat it. You should good it when you encounter it. The answer is controlled by circumstances.
That being noted, b^0 in general is 1.
The y intercept is (0,a)
I believe that the answer is 20.
First point ur point 6,3 and then graph ur slope by going up 1 and right across 4 and down 1 and left across if that’s makes sense
Answer:
answer= -x
Step-by-step explanation:
given: f(x)=3x+4
g(x)=4x+4
f(x)-g(x)= 3x+4-(4x+4)
=3x+4- 4x-4
=3x-4x+4-4
= -x+0
= -x
