The correct question is
Which is the best approximation to a solution of the equation
e^(2x) = 2e^{x) + 3?
we have that
e^(2x) = 2e^{x) + 3-----------> e^(2x)- 2e^{x) - 3=0
the term
e^(2x)- 2e^{x)----------> (e^x)²-2e^(x)*(1)+1²-1²------> (e^x-1)²-1
then
e^(2x)- 2e^{x) - 3=0--------> (e^x-1)²-1-3=0------> (e^x-1)²=4
(e^x-1)=2--------> e^x=3
x*ln(e)=ln(3)---------> x=ln(3)
ln(3)=1.10
hence
x=1.10
the answer is x=1.10
Answer:
-4,-4
Step-by-step explanation:
Answer:
Unit fractions play an important role in modular arithmetic, as they may be used to reduce modular division to the calculation of greatest common divisors. Specifically, suppose that we wish to perform divisions by a value x, modulo y. In order for division by x to be well defined modulo y, x and y must be relatively prime.
Step-by-step explanation:
Answer:
8) No solution
9) n > -5 <em>(View explanation for info on how to graph)</em>
Step-by-step explanation:
<em><u>Question 8</u></em>
<em><u /></em>
<u>24+6k < -6(-4-k)</u> - Solve and graph
1) Distribute -6 to -4 and -k:
24+6k < 24+6k
2) Subtract 6k from both sides:
24<24
^ We know the following is false, therefore there is no solution to question 8.
<em><u>Question 9</u></em>
<em><u /></em>
<u>-2n-40 < 5(6+n)+7n</u> - Solve and graph
1) Distribute 5 to 6 and n:
-2n-40 < 30+5n+7n
2) Combine alike terms:
-2n-40 < 30+12n
3) Add 2n to both sides:
-40 < 30+14n
4) Subtract 30 from both sides:
-70 < 14n
5) Divide both sides by 14:
n > -5
6) To graph, plot an OPEN CIRCLE (<em>don't fill in the circle)</em> at point -5 and then draw an arrow pointing to the right!
Let me know if you need any help with the others :)
Answer:
503.8
Step-by-step explanation:
Since in the hundredth place you have a number higher than 5, we want to round the tenth digit up by one to get 503.8.
