I think this is what you are looking for:
=−12
a
b
=
−
12
+=4
a
+
b
=
4
(+)2=42
(
a
+
b
)
2
=
4
2
2+2+2=16
a
2
+
b
2
+
2
a
b
=
16
∴2+2=16+2×12=40
∴
a
2
+
b
2
=
16
+
2
×
12
=
40
Now, (−)2=2+2−2=40+2×12=64
(
a
−
b
)
2
=
a
2
+
b
2
−
2
a
b
=
40
+
2
×
12
=
64
∴(−)=64‾‾‾√=±8
∴
(
a
−
b
)
=
64
=
±
8
So, 2−2=(+)(−)
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
2−2=(4)(±8)=±32
a
2
−
b
2
=
(
4
)
(
±
8
)
=
±
32
Hope that helps and sorry if it is confusing!
Answer:
4
Step-by-step explanation:
2+2=4
thanks:D
Answer: -1
Step-by-step explanation:
You want to find an angle that is coterminal to 495. So, subtract 360 degrees until youre in the range of 0-360. I got 495 - 360 = 135°
Tangent is equal to
, we already solved theta which was 135°
This next part is hard to explain to someone who doesnt know their trig circle, idk if you do. The angle 135 is apart of the pi/4 gang. So we know this is going to be some variant of √2/2. Sine of quadrant 1 and 2 is gonna be positive:
![sin135=\frac{\sqrt{2} }{2}](https://tex.z-dn.net/?f=sin135%3D%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7D)
Now lets do cosine of 135°, which again is apart of the pi/4 gang because its divisible by 45°. Its in quadrant 2 so the cosine will be negative.
![cos135=-\frac{\sqrt{2} }{2}](https://tex.z-dn.net/?f=cos135%3D-%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7D)
The final step is to divide them. They are both fractions so you should multiply by the reciprocal.
![\frac{\sqrt{2} }{2} *-\frac{2}{\sqrt{2} } =-\frac{2\sqrt{2} }{2\sqrt{2} } =-1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7D%20%2A-%5Cfrac%7B2%7D%7B%5Csqrt%7B2%7D%20%7D%20%3D-%5Cfrac%7B2%5Csqrt%7B2%7D%20%7D%7B2%5Csqrt%7B2%7D%20%7D%20%3D-1)
Well, we can denote L and W for the length and width respectively. Lets say the A is the area, we have: 1. A=(L × W) as well as 2. 2(L+W)=400. We rearrange the second equation to get 3. W=200-L. From this, we can see that 0<L<200. Substitute the third equation into the first to get A=(200L-L²). put this formula into the scientific calculator and you will find a parabola with a maximum. That would be the maximum area of the enclosed area. Alternatively, we can say that L is between 0 and 200 when the area equals 0. (The graph you find will be area against length). As the maximum is generally found halfway, we substitute 100 into the equation and we end up with 10000.
Hope this helps.
The percent of change will always be represented by a positive number because that change is an absolute value. Absolute values will always stay positive because the direction is ignored. It's not like a -1 or a -2, but it's a zero or a 3.
Hope this helped you out! :D