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:
No
Step-by-step explanation:
3:6=1:2 and 6:3=2:1
The frequency of revision times is given by the product of the frequency
density value and the class width.
Correct response:
- The number of students are <u>72 students</u>
<h3>Methods of calculation using a histogram</h3>
The estimate of the number of students who revised for less than 45
minutes is given by the area, <em>A,</em> under the histogram to the left of the 45
minute mark as follows;
Frequency, f = Frequency density × Class width
Number of students = ∑f
Therefore;
A = 5 × 2 + 10 × 2.2 + 20 × 1.6 + 10 × 0.8 = 72
- The number of students that revised for less than 45 minutes = <u>72 </u>students
Learn more about histograms here:
brainly.com/question/17139138
Answer:
|x| + 1
<em>Adding/subtracting on the outside moves it up/down.</em>
<em>Adding/subtracting on the inside (such as (x - 2) moves it right 2) moves it left/right.</em>
<em />
Answer:
t <u>> 3</u>
Step-by-step explanation:
because 16 x 3 = 48 so you can buy 0 - 3 tickets
