You would first have to find common denominators for the fractions. In this case it would be 10. So your new equation would be 3 5/10 - 2 4/10. Once you do this you can then subtract and solve getting the answer of 1 1/10
Answer:
x=9
Step-by-step explanation:
1(x)+0=9
Why?
Because the larger the coeffient, the smaller x gets.
1x=9
x=9
2x=9
x=4.5
3x=9
x=3
You wouldn't want to add any number either, but conviniently, whole numbers also include 0
Answer:
x^2 +y^2
Step-by-step explanation:
Dude this is Pythragorean Theorem =)).
You may have missed this part of your lesson. It's ok and also not ok at the same time beacause pythagorean theorem is a meme.
Ok I'll explain once =)).
Thousands of years ago, a mathematician named Pythagore found out that if we have 3 squares, with one square's area equals to the sum of two other squares, then when we put the sides of the squares together to form a triangle , it'll always be a right triangle ( a triangle with one of its angle is 90°). A square area is measured by the length of its side multipled by itself. So he came up with the statement : In a right triangle, the length of the hypotenuse( the side that does not connect to the right angle) is equalled to the sum of squared other sides.
EG.
In this triangle, the right angle is ^ACB. If length of BC is <em>a</em><em>,</em><em> </em>length of AB is <em>c</em>, length of BC is <em>a</em>
A
I We have a formula :
I \ <em>a</em>^2 + <em>b</em>^2 = <em>c</em>^2
I \
I \
I <em>b</em> \ <em>c</em>
I \
I \
I______\
C <em> a </em> B
So now you can use pythagorean theorem to show off =)).
HOPE YOU LEARN WITH JOY AND HIGH GRADES !!!
Holding what.......? can u be more specific?
First recall that a microsecond is 10^−6 seconds. Hence, one second = 10^6 microseconds, one hour = 3600000000 = 3.6 • 10^9 microseconds, one month (assume a month has 30 days) = 2592000000000 = 2.592 • 10^12 microseconds, and one century = 3110400000000000 = 3.1104 • 10^15 microseconds.
Row 1: f(n) = log n In this case, we need to determine the largest n such that log n ≤ 1000000. To solve this inequality, we need to rewrite the inequality as 2^logn ≤ 2^1000000 or n ≤ 2^1000000. Recall from lecture that 2^10 ≈ 10^3, thus we have that 2^1000000 = 2^10•100000 = (2^10)^100000 ≈ (10^3)^100000 = 10^300000. This is the result given in the textbook. For one hour, we have that n ≤ 2^3600•1000000 and thus n ≈ 10^1080000000.
Row 8: f(n) = n! For us to see that the largest sized input is 12 that can be processed within an hour when f(n) = n! one can simply, compute 12! And verify that it is less than the number of microseconds in one hour, but that 13! is greater than the number of microseconds in an hour.
Row 4: f(n) = n log n In this case, use Maple to solve equations like n log n−1000000 = 0. The Maple command for solving this equation is fsolve(n*log[2](n) - 1000000 = 0)
