Answer:
2 cis (7/6 pi)
Step-by-step explanation:
r = sqrt( a^2 + b^2)
r = sqrt (-sqrt(3) )^2 + (-1)^2)
= sqrt(3 +1)
= sqrt(4)
= 2
theta = arctan (b/a)
theta = arctan (-1/-sqrt(3))
theta = 30
but this is in the third quardrant -a and -b
so add 180
theta = 210 degrees
convert this to radians
210 * pi/180 = 210/180 * pi = 21/18 * pi = 7/6 * pi
r cis (theta)
2 cis (7/6 pi)
Answer:
45/4
Step-by-step explanation:
We can interpret the question to have two equation which can be solve simultaneously
a+b=7------------(1)
a-b=2------------(2)
From eqn(2) make a subject of formula
a=2+b--------(3)
Substitute the (3) into eqn(1)
a+b=7------------(1)
2+b+b=7
2b=7-2
2b=5
b=5/2
From equation (3) substitute value of b to find a
a=2+b--------(3)
a= 2+5/2
a=9/2
Then What is the value of a x b ?
9/2× 5/2
=45/4
Quick answer I don't think this has an answer.
If you take the cos-1(2 sqrt(2)) your calculator should have a fit. Let's check that out. Mine certainly does. So there is something wrong with the question. If there is something to add in please do it and I will it least put an answer in the comments. As it stands, nothing will work.
If you put your calculator in radians, you will get an answer but it will not be anything resembling the choices you've listed.
If you meant sqrt(2) / 2 that would give 45o. Put it in your calculator like this 2 ^ 0.5 divided by 2 = 0.707
Cos - 1 (0.707) = 45
Answer:
There may be 1 or 3 tricycles in the parking lot.
Step-by-step explanation:
Since at any point in time, there could be bicycles, tricycles, and cars in the school parking lot, and today, there are 53 wheels in total, if there are 15 bicycles, tricycles, and cars in total, to determine how many tricycles could be in the parking lot, the following calculation must be performed:
13 x 4 + 1 x 3 + 1 x 2 = 57
11 x 4 + 1 x 3 + 3 x 2 = 53
10 x 4 + 3 x 3 + 2 x 2 = 53
8 x 4 + 5 x 3 + 2 x 2 = 51
10 x 2 + 1 x 3 + 4 x 4 = 39
9 x 3 + 1 x 2 + 5 x 4 = 49
Therefore, there may be 1 or 3 tricycles in the parking lot.
No no yes yes
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