Answer:
f(x)=-18x^2
Step-by-step explanation:
Given:
1+Integral(f(t)/t^6, t=a..x)=6x^-3
Let's get rid of integral by differentiating both sides.
Using fundamental of calculus and power rule(integration):
0+f(x)/x^6=-18x^-4
Additive Identity property applied:
f(x)/x^6=-18x^-4
Multiply both sides by x^6:
f(x)=-18x^-4×x^6
Power rule (exponents) applied"
f(x)=-18x^2
Check:
1+Integral(-18t^2/t^6, t=a..x)=6x^-3
1+Integral(-18t^-4, t=a..x)=6x^-3
1+(-18t^-3/-3, t=a..x)=6x^-3
1+(6t^-3, t=a..x)=6x^-3
That looks great since those powers are the same on both side after integration.
Plug in limits:
1+(6x^-3-6a^-3)=6x^-3
We need 1-6a^-3=0 so that the equation holds true for all x.
Subtract 1 on both sides:
-6a^-3=-1
Divide both sides by-6:
a^-3=1/6
Raise both sides to -1/3 power:
a=(1/6)^(-1/3)
Negative exponent just refers to reciprocal of our base:
a=6^(1/3)
Answer:
13
Step-by-step explanation:
Answer:
Step-by-step explanation:
In propositional logic and Boolean algebra, De Morgan's laws are a pair of transformation rules that are both valid rules of inference. ... The rules allow the expression of conjunctions and disjunctions purely in terms of each other via negation.
Answer:
y = 3x-16
Step-by-step explanation:
two points where the line goes through is (5,-4) as mentioned and (0,-16)
Answer:
EG = 16 and FH =22
Step-by-step explanation:
We know that the diagonals of a parallelogram bisect each other
so 2a = 3b+2
and 2a+3 = 6b-1
We know have a system of equations to solve
2a = 3b+2
2a+3 = 6b-1
Subtract 3 from each side
2a+3-3 = 6b-1-3
2a = 6b -4
Now we can set the 2 equations equal ( 2a = 3b+2 and 2a = 6b -4)
3b+2 = 6b-4
Subtract 3b from each side
3b-3b+2 = 6b-3b-4
2 = 3b-4
Add 4 to each side
2+4 = 3b-4+4
6 = 3b
Divide by 3
6/3 = 3b/3
2 =b
We want to find a
2a = 3b+2
Substitute in b=2
2a = 3(2) + 2
2a = 6+2
2a =8
Divide by 2
2a/2 =8/2
a = 4
Now that we know a and b
EG = 2a + 3b+2
= 2(4) + 3(2)+2
= 8+6+2
= 16
FH = 2a+3 + 6b-1
= 2(4) +3 +6(2)-1
= 8+3+12-1
= 23-1
= 22
