Answer:
No Real Solutions.
Step-by-step explanation:
√(x-5)=8√(x-2)
Square on both sides,
(√(x-5))^2=(8√(x-2))^2
or,(x-5)=64(x-2)
or,x-5=64x-128
or,-5+128=64x-x
or,123/63=x
Therefore, x=41/21
If you plug x=41/21 in √(x-5)=8√(x-2) you do not get equals, so the problem has no real solutions.
Answer:
2
Step-by-step explanation:
Given g(x) = sin(x)-1/cos2(x), we are to find the limit if the function g(x) as g(x) tends to π/2
Substituting π/2 into the function
lim x-->π/2 sin(x)-1/cos 2(x)
= sin(π/2) - 1/cos(2)(π/2)
= 1 - 1/cosπ
= 1- 1/-1
= 1 -(-1)
= 1+1
= 2
Hence the limit of the function h(x) = sin(x)-1/cos2(x) as x--> π/2 is 2
Answer:
x= 1/e or x= 0.367
Step-by-step explanation:
:))) Your welcome
Answer:
Step-by-step explanation:
The time intervals are [0.00, 0.10], [0.10, 0.30], and [0.30, 0.60].
The rate of change in each interval is the slope of the line between the endpoints:
m = [f(b) − f(a)] / (b − a)
Filling out the table:
![\left[\begin{array}{ccc}Time (s)&Height (ft)&Rate\ of\ change(ft/s)\\0.00&96.44\\&&\frac{97.60-96.44}{0.10-0.00}=11.6 \\0.10&97.60\\&&\frac{98.57-97.60}{0.30-0.10}=4.85\\0.30&98.57\\&&\frac{98.30-98.57}{0.60-0.30}=-0.90\\0.60&98.30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DTime%20%28s%29%26Height%20%28ft%29%26Rate%5C%20of%5C%20change%28ft%2Fs%29%5C%5C0.00%2696.44%5C%5C%26%26%5Cfrac%7B97.60-96.44%7D%7B0.10-0.00%7D%3D11.6%20%5C%5C0.10%2697.60%5C%5C%26%26%5Cfrac%7B98.57-97.60%7D%7B0.30-0.10%7D%3D4.85%5C%5C0.30%2698.57%5C%5C%26%26%5Cfrac%7B98.30-98.57%7D%7B0.60-0.30%7D%3D-0.90%5C%5C0.60%2698.30%5Cend%7Barray%7D%5Cright%5D)
The intervals are already in order from greatest to least. Simply reverse the order to get least to greatest:
[0.30, 0.60]
[0.10, 0.30]
[0.00, 0.10]