Maurice wants to create a set of elliptical flower beds. To do this, he first plots the location of the two fruit trees on his graph.
Maurice has to use the equation a^2-b^2=c^2. We know that c=3, and because we need 1 more number to solve for b, I made a=6. 6^2-b^2=3^2. 36-b^2=9. b^2=27. b=5.196
<span>Next, to create the equation, we substitute what we know into the equation x^2/a^2 + y^2/b^2=1 and get x^2/36 + y^2/27=1. Johanna wants to create some hyperbolic flower beds.
We already know that c=3 so this time I decided a=1. 3^2=1^2+b^2. 9=1+b^2. 8=b^2. b=2.828
Next, to create the equation, we substitute what we know to the equation x^2/a^2 - y^2/b^2 = 1. x^2/1^2 - y^2/2.828^2 = 1. </span>
(a+bi)(2+3i)-i=(-11+5i)(1-i)
2a+3ai+2bi+3bi²-i=-11+11i+5i-5i²
2a+(3a+2b-1)i-3b=-11+16i+5
(2a-3b)+(3a+2b-1)i=-6+16i
Therefore; we have the next system of equations:
2a-3b=-6
3a+2b-1=16
Finally, the system of equations is:
2a-3b=-6
3a+2b=17
We can solve this system of equations by reduction method
2(2a-3b=-6)
3(3a+2b=17)
----------------------
13a=39 ⇒a=39/13=3
3(2a-3b=-6)
-2(3a+2b=17)
----------------------
-13b=-52 ⇒ b=52/13=4
Answer: a=3; b=4
Answer:
m
parallel
=
−
1
3
Step-by-step explanation:
Answer:
12
Step-by-step explanatik just need the points
8y-1-28y-20-6y+27
8y-28y-6y-1-20+27
-26y+6