Answer:
the answer is 19
Step-by-step explanation:
Answer:
the linear function gives us the points
(0,3)
(4,0)
(8,-3)
find the equation
hm
find slope
(y2-y1)/(x2-x1)
for (0,3) and (4,0)
slope=(0-3)/(4-0)=-3/4
y=-3/4x+b
given
(0,3)
y=-3/4x+3
so sub -3/4x+3 for y in other equation
(-3/4x+3)^2+x^2=4
expand
9/2x^2-9/2x+9+x^2=4
11/2x^2-9/2x+9=4
times 2 both sides
11x^2-9x+18=8
minus 8 both sides
11x^2-9x+10=0
factor
or use quadratic formula
for
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}
2a
−b+/−
b
2
−4ac
11x^2-9x+10=0
x=\frac{-(-9)+/- \sqrt{(-9)^2-4(11)(10)} }{2(11)}
2(11)
−(−9)+/−
(−9)
2
−4(11)(10)
x=\frac{9+/- \sqrt{81-440} }{22}
22
9+/−
81−440
x=\frac{9+/- \sqrt{-359} }{22}
22
9+/−
−359
we see we have no real roots
there are no intersection points
#Carry on learning
#Correct me if I'm wrong
Answer:
yes
Step-by-step explanation:
If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi. Imaginary solutions always come in pairs.
So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.
(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.
(x-2i)(x+2i)(x-5i)(x+5i) = (x^2+4)(x^2+25)
= x^4 + 29x^2 + 100
So the equation would be x^4 + 29x^2 + 100 = 0.
Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.
Answer:
8,874%
Step-by-step explanation:
Find percent of change by finding the differnce between the original price and new price. You can then use a proportion "change/original = percent/100"
