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:
Step-by-step explanation:
<u>Use formula for continuous interest:</u>
- A =
, where r- interest rate, t- time in years
<u>If A = 3P, and r = 0.085 then the equation is:</u>
- ln 3 = 0.085t
- 1.099 = 0.085t
- t = 1.099 / 0.085
- t = 12.9 ≈ 13 years
Answer:
what is this
Step-by-step explanation:
The roots of the polynomial <span><span>x^3 </span>− 2<span>x^2 </span>− 4x + 2</span> are:
<span><span>x1 </span>= 0.42801</span>
<span><span>x2 </span>= −1.51414</span>
<span><span>x3 </span>= 3.08613</span>
x1 and x2 are in the desired interval [-2, 2]
f'(x) = 3x^2 - 4x - 4
so we have:
3x^2 - 4x - 4 = 0
<span>x = ( 4 +- </span><span>√(16 + 48) </span>)/6
x_1 = -4/6 = -0.66
x_ 2 = 2
According to Rolle's theorem, we have one point in between:
x1 = 0.42801 and x2 = −1.51414
where f'(x) = 0, and that is <span>x_1 = -0.66</span>
so we see that Rolle's theorem holds in our function.
Answer:
v=5.42m/s
Step-by-Step Explanation:
We can use conservation of energy to solve this
mgh=1/2mv^2
2gh=v^2
v=sqrt(2gh)
v=sqrt[2*9.8m/s^2*(6.5-5)]
v=5.42m/s
