can you assume that any fraction is in simplest form if either the numerator or denominator is a prime number?
2 answers:
You might be interested in
We can first add up the cards so we know how many we have in all:
16 + 16 + 18 = 50 cards
We can do this a little bit easier if we get the "16"-cards in one number total.
16 + 16 = 32
= 32 x 2 = 
50 x 2
= 64 : 2 = 32 %
100
We did just divide the % of two types cards on 2, so we get the %-chance of 1 type card.
I am not quite sure, but I think that 32 % is the correct answer.
16 + 16 + 18 = 50 cards
We can do this a little bit easier if we get the "16"-cards in one number total.
16 + 16 = 32
50 x 2
100
We did just divide the % of two types cards on 2, so we get the %-chance of 1 type card.
I am not quite sure, but I think that 32 % is the correct answer.
I would compute sqrt(1 + 160pi^2) first to get approximately 39.75093337
Add this to 1 and we have 40.75093337
Then divide over 2pi to get a final approximate result of 6.48571248
So x = 6.48571248 is one approximate solution
In short, I computed only focusing on the plus for now.
----------------------
If you were to compute you should get roughly -6.167402596 as your other solution. Each solution can then be plugged into the original equation to check if you get 0 or not. You likely won't land exactly on 0 but you'll get close enough.
Answer:
a rhombus
Step-by-step explanation:
If you graph the problem, you can see the shape of the quadrilateral. I attached a picture of a graph below. I hope this helped you!! Have a great rest of your day.
