Simplifying the given expressions we proceed as follows:
(5sqrt3)^x
=5^x*(3^1/2)^x
=5^x*3^x/2
=5^x3^u
where u=x/2
(1/2)^(x-3)
=1/2^(x-3)
=2^-(3-x)
=2^u
where u=-(3-x)
9/3sqrt(3)
=3/(3)^(1/2)
=3(3)^(-1/2)
16/(3sqrt (2^x))
=1/3*(2^4*2^(-x/2))
=1/3*2^(4-x/2)
=1/3*2^u
where:
u=4-x/2
Answer:
Step-by-step explanation:
given is the Differential equation in I order linear as
Take Laplace on both sides
![L(y') +4L(y) = 48L(t)\\sY(s)-y(0) +4Y(s) = 48 *\frac{1}{s^2} \\Y(s) [s+4]=\frac{48}{s^2}+9\\Y(s) = \frac{1}{s^2(s+4)}+\frac{9}{s+4}](https://tex.z-dn.net/?f=L%28y%27%29%20%2B4L%28y%29%20%3D%2048L%28t%29%5C%5CsY%28s%29-y%280%29%20%2B4Y%28s%29%20%3D%2048%20%2A%5Cfrac%7B1%7D%7Bs%5E2%7D%20%5C%5CY%28s%29%20%5Bs%2B4%5D%3D%5Cfrac%7B48%7D%7Bs%5E2%7D%2B9%5C%5CY%28s%29%20%3D%20%5Cfrac%7B1%7D%7Bs%5E2%28s%2B4%29%7D%2B%5Cfrac%7B9%7D%7Bs%2B4%7D)
Now if we take inverse we get y(t) the solution
Thus the algebraic equation would be
Answer:
Sue's scores for the four games in ascending order are: 97, 98, 98, 107
Step-by-step explanation:
Her modal score was 98. The mode is found by using the number that appears most often. This means that 98 has to appear at least two times out of the four scores.
Her range was 10. The range is found by taking the highest score and subtracting it from the lowest score. The highest score had to be greater than 98 and the lowest score had to be less than 98 since we know the mode was 98.
Her mean score was 100. This mean is found by adding all the numbers together and then dividing by the total numbers listed. Adding the four scores together and dividing by 4 will equal 100.
Used guess and test:
Highest Number, 98, 98, Lowest Number
107 - 97 = 10 (meets range requirement)
97 + 98 + 98 + 107 = 400
400/4 = 100 (meets the mean requirement)
