WILL MARK BRAINLIEST!! Solve for k. -5 < -5k - 15 < 20
1 answer:
You might be interested in
a(i). Since you are given a velocity v. time graph, the distance will be represented by:
In this case, however, we can just use simple geometry to evaluate the area under the graph v(t). I split it up into 2 trapezoids, and 1 rectangle. So, the area will be as follows:
So, the particle traveled a total of 1275m assuming it never turned back (because it says to calculate distance).
a(iii). Deceleration is a word for negative acceleration. Acceleration is the first derivative of velocity, and so deceleration is too. So, we just need to find the slope of the line that passes through t = 30 because it has a linear slope (meaning the slope doesn't change). So, we can just use simple algebra instead of calculus to figure this out. Recall from algebra that slope (m):
So, let's just pick values. I'm going to pick (25, 30) and (35, 15). Let's plug and chug:
Since it's a negative value, this means that acceleration is negative but deceleration is positive (because deceleration is negative acceleration). So, your answer is: The deceleration of the particle at t = 30s is 3/2 or 1.5.
In this case, however, we can just use simple geometry to evaluate the area under the graph v(t). I split it up into 2 trapezoids, and 1 rectangle. So, the area will be as follows:
So, the particle traveled a total of 1275m assuming it never turned back (because it says to calculate distance).
a(iii). Deceleration is a word for negative acceleration. Acceleration is the first derivative of velocity, and so deceleration is too. So, we just need to find the slope of the line that passes through t = 30 because it has a linear slope (meaning the slope doesn't change). So, we can just use simple algebra instead of calculus to figure this out. Recall from algebra that slope (m):
So, let's just pick values. I'm going to pick (25, 30) and (35, 15). Let's plug and chug:
Since it's a negative value, this means that acceleration is negative but deceleration is positive (because deceleration is negative acceleration). So, your answer is: The deceleration of the particle at t = 30s is 3/2 or 1.5.
9514 1404 393
Answer:
the cubic does not exceed the quadratic until x > 8
Step-by-step explanation:
The attached graph plots the two expressions. The two are equal at x=8. Below that value the expression 7x^2 is greater. Above that value, the expression 7x^3 -x^2 is greater.
_____
<em>Comment on the question</em>
The directive to "compare the volumes" does not tell anything about the sort of comparison you are expecting. We note that the expression x^3-x^2 is zero or less up to the point where x=1, so we wonder exactly what it is supposed to be modeling. Any real volume will not be negative.
For small values of x, the quadratic is quite a bit larger. However, we know the cubic will grow larger than any quadratic for large values of x. So, the comparison you get will depend on the domain of interest—which is not specified.
