Answer:
x^2 + 2xy + y ^2
Step-by-step explanation:
A = a^2
a = x + y
(x + y) + (x + y)
x^2 + xy + xy + y^2
x^2 + 2xy + y ^2
Short AnswerThere are two numbers
x1 = -0.25 + 0.9682i <<<<
answer 1x2 = - 0.25 - 0.9582i <<<<
answer 2 I take it there are two such numbers.
Let one number = x
Let one number = y
x + y = -0.5
y = - 0.5 - x (1)
xy = 1 (2)
Put equation 1 into equation 2
xy = 1
x(-0.5 - x) = 1
-0.5x - x^2 = 1 Subtract 1 from both sides.
-0.5x - x^2 - 1 = 0 Order these by powers
-x^2 - 0.5x -1 = 0 Multiply though by - 1
x^2 + 0.5x + 1 = 0 Use the quadratic formula to solve this.

a = 1
b = 0.5
c = 1

x = [-0.5 +/- sqrt(0.25 - 4)] / 2
x = [-0.5 +/- sqrt(-3.75)] / 2
x = [-0.25 +/- 0.9682i
x1 = -0.25 + 0.9682 i
x2 = -0.25 - 0.9682 i
These two are conjugates. They will add as x1 + x2 = -0.25 - 0.25 = - 0.50.
The complex parts cancel out. Getting them to multiply to 1 will be a little more difficult. I'll do that under the check.
Check(-0.25 - 0.9682i)(-0.25 + 0.9682i)
Use FOIL
F:-0.25 * -0.25 = 0.0625
O: -0.25*0.9682i
I: +0.25*0.9682i
L: -0.9682i*0.9682i = - 0.9375 i^2 = 0.9375
NoticeThe two middle terms (labled "O" and "I" ) cancel out. They are of opposite signs.
The final result is 0.9375 and 0.0625 add up to 1
Step-by-step explanation:
∫ dt / (cos²(t) ⁹√(1 + tan(t)))
If u = 1 + tan(t), then du = sec²(t) dt.
∫ du / ⁹√u
∫ u^(-1/9) du
9/8 u^(8/9) + C
9/8 (1 + tan(t))^(8/9) + C
So here we use the pythagorean theorem which is a2 + b2 = c2 (“a” squared times “b” squared equals “c” squared) the length from calvins house to the intersection is “a” and the length from phoebes house to the intersection is “b” so in order to find out the length of “c” (calvins house to phoebes house) we need to use the pythagorean theorem
a2 275x275=75,625
b2 113x113 = 12,769
so now that we have figured out what a2 and b2 are let’s add them
75,625 + 12,769 = 88,394
now since we only need to find out the length and not the area we need to find the square root of 88,394
the square root of 88,394 is 297.311 (i cut off the decimal after three places)
the direct length from phoebes house to calvins is 297.311 meters
Answer:
1) C. (3, 5)
2) B. Counterclockwise turn 90 degrees about the origin
3) A. C' (0, -2)
Step-by-step explanation:
i got u bro