What is the answer to... 5m+8n-2(3m-n)? Show step by step
1 answer:
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Answer:
Detailed solution is given below:
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.
Answer:
Step-by-step explanation:
the word ''to'' means over
So, from 1992 to 1997, it went up by some "r" rate, ok.. that means some percentage, that means some rate of growth, so is an exponential function, with a positive rate, or +r
if we take 1992, to be 0years, then the starting amount for the tuition is 1685
that is
now, let's go to 1997, 5 years later, when t = 5, we know the tuition price then was 2392, so A = 2392
thus![\bf A=P\left(1+r\right)^t \quad \begin{cases} A=\textit{accumulated amount}\to &\$2392\\ P=\textit{starting amount}\to &\$1685\\ r=rate\\ t=years\to &5 \end{cases} \\\\\\ 2392=1685(1+r)^5\implies \cfrac{2392}{1685}=(1+r)^5\implies \sqrt[5]{\cfrac{2392}{1685}}=1+r \\\\\\ \boxed{\sqrt[5]{\cfrac{2392}{1685}}-1=r}](https://tex.z-dn.net/?f=%5Cbf%20A%3DP%5Cleft%281%2Br%5Cright%29%5Et%0A%5Cquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5Cto%20%26%5C%242392%5C%5C%0AP%3D%5Ctextit%7Bstarting%20amount%7D%5Cto%20%26%5C%241685%5C%5C%0Ar%3Drate%5C%5C%0At%3Dyears%5Cto%20%265%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A2392%3D1685%281%2Br%29%5E5%5Cimplies%20%5Ccfrac%7B2392%7D%7B1685%7D%3D%281%2Br%29%5E5%5Cimplies%20%5Csqrt%5B5%5D%7B%5Ccfrac%7B2392%7D%7B1685%7D%7D%3D1%2Br%0A%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7B%5Csqrt%5B5%5D%7B%5Ccfrac%7B2392%7D%7B1685%7D%7D-1%3Dr%7D)
now, you'd get a value in decimal format, so, to get the % format, simply multiply it by 100
if we take 1992, to be 0years, then the starting amount for the tuition is 1685
that is
now, let's go to 1997, 5 years later, when t = 5, we know the tuition price then was 2392, so A = 2392
thus
now, you'd get a value in decimal format, so, to get the % format, simply multiply it by 100