Solution
For this case we can take square root in both sides and we have:
![3x-5=\pm\sqrt[]{19}](https://tex.z-dn.net/?f=3x-5%3D%5Cpm%5Csqrt%5B%5D%7B19%7D)
And solving for x we got:
![x=\frac{5\pm\sqrt[]{19}}{3}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5%5Cpm%5Csqrt%5B%5D%7B19%7D%7D%7B3%7D)
then the solutions for this case are:
B and E
We'll assume this is an arbitrary triangle ABC.
A) No, the sines of two different angles can be whatever they want
B) sin(B)=cos(90-B)
Yes, that's always true. The "co" in cosine means "complementary" as in the complementary angle, which adds to 90. So the sine of an angle is the cosine of the complementary angle.
C) No, the correct identity is sin(180-B)=sin B. Supplementary angles share the same sine.
D) Just like A, different triangle angles often have different cosines.
Answer: Choice B
slope = (25-4)/(30-10)
slope = 21/20
slope =21/20
using point slope form
y-y1 = m(x-x1)
y-4 = 21/20 (x-10)
y = 21/20x -21/2 +4
y = 21/20 x -21/2 +8/2
y = 21/20x -13/2
let x=40
y = 21/20 (40) -13/2
y = 42-13/2
y = 35.5 games
If we round the slope to 1
slope =1
using point slope form
y-y1 = m(x-x1)
y-4 = 1 (x-10)
y = 1x -10 +4
y = x -6
let x=40
y = 40-6
y = 34 games
Answer:
4 I believe
Step-by-step explanation:
sorry if im wrong, (didn't know if the and was to add or multiply,sorry!)
Answer:
-3
Step-by-step explanation:
