I don’t want you going on a trip with x+2-1 coordinate of Q
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
Here you go:
h = 5 x 2.54
please correct me if I'm wrong,
Hope it helps!
The answer to your question is <u>3</u>
1/2, because 3/10 + 1/5 = 5/10. that means that the other part of 1 is 5/10 or 1/2
