I think just two reflections would do it.
First we reflect around y = -x, the 45 degree line through the origin and the second and fourth quadrant.
Then we reflect through the y axis, x=0.
The composition of the two reflections is equivalent to a 90 degree clockwise rotation.
<span>The discriminant of a quadratic equation is the b^2-4ac portion that the square root is taken of. If the discriminant is negative, then the function has 2 imaginary roots, if the discriminant is equal to 0, then the function has only 1 real root, and finally, if the discriminant is greater than 0, the function has 2 real roots. So let's look at the equations and see which have a positive discriminant.
f(x) = x^2 + 6x + 8
6^2 - 4*1*8
36 - 32 = 4
Positive, so f(x) has 2 real roots.
g(x) = x^2 + 4x + 8
4^2 - 4*1*8
16 - 32 = -16
Negative, so g(x) does not have any real roots
h(x) = x^2 – 12x + 32
-12^2 - 4*1*32
144 - 128 = 16
Positive, so h(x) has 2 real roots.
k(x) = x^2 + 4x – 1
4^2 - 4*1*(-1)
16 - (-4) = 20
Positive, so k(x) has 2 real roots.
p(x) = 5x^2 + 5x + 4
5^2 - 4*5*4
25 - 80 = -55
Negative, so p(x) does not have any real roots
t(x) = x^2 – 2x – 15
-2^2 - 4*1*(-15)
4 - (-60) = 64
Positive, so t(x) has 2 real roots.</span>
Answer:
58, 37, 9
Step-by-step explanation:
Given:
First term a₁ = 65 and common difference d = - 7
This sequence is arithmetic series and formula for calculating n-th term is:
aₙ = a₁ + (n-1) d
Accordingly
The second term is:
a₂ = 65 + (2-1) (-7) = 65 - 7 = 58
a₂ = 58
The fifth term is:
a₅ = 65 + (5 - 1) (-7) = 65 + 4 · (-7) = 65 - 28 = 37
a₅ = 37
The ninth term is:
a₉ = 65 + (9 - 1) (-7) = 65 + 8 · (-7) = 65 - 56 = 9
a₉ = 9
God with you!!!
Answer:
AB=5.4 units
Step-by-step explanation:
We using the theorem of intersecting tangent and secant to solve this.
By this theorem:
