First, find the number of inches
<span>1225 Miles = 77616000 Inches
then, move the decimal to until you have a ones place. </span>
<span>77616000. the decimal for this number moves to the left SEVEN places.
you end up with 7.7616000
</span>
it will look like this:
7.8 * 10^7
I rounded to the .8 because you didn't state how many significant figures, but it could easily be 7.76 * 10^7
the exponent above the 10 will tell you how many places to move. to get back to your original number, move it back to the right seven places. Of course, you would have to place hold the numbers with 0s. If it were negative, you would move the decimal the opposite way.
Answer: 89.6 kg of water
Step-by-step explanation:
Find the total volume of the tank:
(80cm)*(35cm)*(40cm) = 112,000 cm^3
4/5 of this would be 89,600 cm^3
Water has a mass of 1 gram/cm^3
(1 gram/cm^3)*(89,600 cm^3) = 89,600 g of water
(89,600 g)*(1 kg/1,000 g) = 89.6 kg of water
Imaginary number is one that when squared gives a negative result. ... With imaginary numbers, when you square them, the answer is negative. They are written like a real number, but with the letter i after them, like this: 23iThe letter i means it is an imaginary number.
Answer:
The average rate of change of the given function
A(x) = 1
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given function f(x) = x² - 2x -4
And given that x = a = -1 and x=b = 4
The average rate of change of the given function
<u><em>Step(ii):-</em></u>
f(x) = x² - 2x -4
f(-1) = (-1)² - 2(-1) -4 = 1+2-4 = -1
f(4) = 4² -2(4) -4 = 16 -12 = 4
The average rate of change of the given function
<u><em>final answer:-</em></u>
The average rate of change of the given function
A(x) = 1
Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
