Answer:
t = [0:0.01:4];
x = 5*t - 10;
y = 25442 - 120*t + 144;
d = x.^2 + y.^2;
min = 1e+14;
for k = 1:length(t)
if d(k) < min
min = da(k);
tmin = t(k);
end
end
disp('The minimum distance is: ')
disp(sqrt(min))
disp('and it occurs at t = ')
disp(tmin)Output:
>> withLoop
The minimum distance is:
2.5106e+04
and it occurs at t =
4
Explanation:
Answer:








Explanation:
Solving (a): To base 10

We simply multiply each digit by a base of 16 to the power of their position.
i.e.


In hexadecimal

So:



This gives:


In hexadecimal

So:


Solving (b): To base 10

We simply multiply each digit by a base of 2 to the power of their position.
i.e.





Solving (c): To base 16

First, convert to base 10
In (b)

Next, is to divide 781754 by 16 and keep track of the remainder





Write out the remainder from bottom to top

In hexadecimal


So:


In b

Next, is to divide 11057389 by 16 and keep track of the remainder






Write out the remainder from bottom to top

In hexadecimal


So:

Solving (d): To octal

Divide 74510 by 8 and keep track of the remainder






Write out the remainder from bottom to top


Divide 67210 by 8 and keep track of the remainder






Write out the remainder from bottom to top

Answer:
The default margins for Microsoft Word from version 2007 onward have been 1 inch (25.4 mm) all around; in Word 2003, the default top and bottom margins were 1 inch (25.4 mm), but 1.25 inches (31.7 mm) were given at the left and the right.
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