Hi there, (6z²+z-1)(9z-5)=(6z²+z+-1)(9z+-5)=(6z²)(9z)+(6z²)(-5)+(z)(9z)+(z)(-5)+(-1)(9z)+(-1)(-5)=54z³-30z²+9z²-5z-9z+5=54z³-21z²-14z+5. Therefore, the answer is 54z³-21z²-14z+5.
Answer:
Step-by-step explanation:
In Δ PQS and ΔPQT
∠QPS = ∠RPT {Common angle}
∠PSQ = ∠PTR = 90° {QS & RT are altitude}
QS = RT {given}
ΔPQS ≅ ΔPRT {A A S congruent}
PQ = PS {CPCT}
So, PQR is an isosceles triangle.
9)
a) Flas card (I) and (iii) are congruent.
b) A S A congruent
c) BC ≅ RP
Answer:
B
Step-by-step explanation:
That's a huge number of points.
First: Calculate the value of 62 two pointers
62 * 2 = 124
Second: find the number of points that were due to 3 pointers.
151 - 124 = 27
Third: Divide by 3 to find the number of 3 pointers.
27÷3 = 9
Answer: There were 9 three pointers.
F(x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.
g(x) = 2x^2-9 is a parabola having vertex at (0, -9) and opening upwards.
By symmetry, let the x-coordinates of the vertices of rectangle be x and -x => its width is 2x.
Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.
=> Area, A
= 2x (y1 - y2)
= 2x (18 - x^2 - 2x^2 + 9)
= 2x (27 - 3x^2)
= 54x - 6x^3
For area to be maximum, dA/dx = 0 and d²A/dx² < 0
=> 54 - 18x^2 = 0
=> x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)
d²A/dx² = - 36x < 0 for x = √3
=> maximum area
= 54(√3) - 6(√3)^3
= 54√3 - 18√3
= 36√3.
