ByVic214
If you have a fraction that contains, for example, a/b*a, that is b because a in the numerator and a in the denominator cancel each other out
If I now have a^5/a^6, that would be a, right? Because you subtract the exponents during division, and a^1 is simply a.
I just want to be sure and I don't rely too much on math AIs
LG
(1 votes)
Loading...Hello, can you send me a good video on how I learn to calculate expressions? I can't seem to find a good one, and does the system of equations (image) apply to both the expression and the system of equations? Could you please also send me a good video for the system of equations? Thanks…
Hello dear math experts, In preparation for an upcoming math exam, I have done some practice exercises and am not sure about this one: For b) I obtained the following values by inserting the 3 side lengths into the corresponding cosine formula. Alpha: 25.7 degrees Beta: 40.77 degrees Gamma: 113.53 degrees But aren't the interior…
I want to create my own math library in Python and am currently working on sine. I'm using the unit circle as a guide. I'm creating a right-angled triangle within it. By entering a second angle, I can determine the third angle. I also know the length of the hypotenuse (c=1). How can I calculate…
2x * sin(x) Can this be simplified
Can someone help me with a and b
In 11a) true, 11b) also support vector important, 11c) true
to remain
a/(ba) = 1/b
yes, a truncated
.
^5/a^6 is 1/a^1 = 1/a
.
The rule is both
subtracting the exponents (starting from the counter)
=
^(1-1)/b^1 =
=
1/b
the something high zero is always 1!
.
^(5-6) = a^-1 and a^-1 = 1/a^1 = 1/a
No. a/(b*a) is 1/b, (a/b)*a is a2/b
This is
a.
a/(b*a)=1/b
^5/a^6=a^(-1)=1/a
The second fracture is 1/a, because 5-6=1 and
^(-1) = 1/a