Answer:
what do u need help with
Step-by-step explanation:
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First step: partition the number you want to square root into a block of 2 digits, starting from the last digit (first diagram)
Second step: As our number is a five-digits, we ends up with 2 28 01. Pick a number that could be squared to get the first partition, 2. This number is 1, since 1×1=1
Third step: Write 1 on the top and on the side, as shown in the second graph
Fourth step: Double the number on the side, which is 1+1=2 and use this number as the first digit for the next multiplier. Meanwhile, subtract 1 from 2 inside the root sign to get 1, then pull the other two digits, 28
Fifth step: We need a value in the boxes that when we multiply together will give a number less than 128. We choose 5 as 25×5=125
Sixth step: Subtract 125 from 128 to give 3, and as the same concept with long division, bring down the 0 and the 1. So we have 301
Seventh step: Add 5 to the multiplier on the left, so 20+5=30, which we will use on the side as the hundred and ten digits.
Final step: Find a number to fit in the boxes. We choose 1 since 301×1=301
And hence the square root of 228801 is 151
Option B
The amount of mice when t = 0 is 300
<h3><u>Solution:</u></h3>
Given that population of mice grows according to the formula:

where t is measured in months
<em><u>To find: amount of mice when t = 0</u></em>
So we have to substitute t = 0 in given formula and find value of A(t)

Substitute t = 0


So we get,

Thus the amount of mice when t = 0 is 300
Answer: He drove 139 miles.
Step-by-step explanation:
0.98x + 20.95= 157.17 where x is the number of miles driven.
-20.95 =-20.95
0.98x = 136.22
x= 139
Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.
At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.
We parameterize the line segment by


with

. Then we find the differential:


with

.
Here, the line integral is





as required.