let's keep in mind that, in the II Quadrant, cosine is negative and sine, is positive.
cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.
Answer:
Step-by-step explanation:
When the coefficients don't lend themselves to solution by substitution or elimination, then Cramer's Rule can be useful. It tells you the solutions to
are ...
Using that rule here, we find ...
∆ = 5·3 -6·2 = 3
a = (5·54 -6·41)/3 = 5·18 -2·41 = 90 -82 = 8
s = (41·3 -54·2)/3 = 41 -18·2 = 5
This math can be performed in your head, which is the intent of formulating the rule in this way.
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Similarly, if you expect the solutions to be small integers (as here), then graphing is another viable solution method.
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<em>Comment on the question</em>
We're sad to see than only 16 tickets were sold to the two performances by the symphonic band.
Answer:
− 19x + 12
Step-by-step explanation:
Answer:
Length of each side of the square (s) = 8 cm
Area of the square (a) = ?
✐ Area of the square is <u>6</u><u>4</u><u> </u><u>square</u><u> </u><u>centimetres.</u>
ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ
# ꧁❣ RainbowSalt2²2² ࿐
