<span>Difference of squares method is a method that is used to evaluate the difference between two perfect squares.
For example, given an algebraic expression in the form:
can be factored as follows:
From the given expressions, the only expression containing two perfect squares with the minus sign in the middle is the expression in option A.
i.e.
which can be factored as follows:
.</span>
Answer:
9-63x
Step-by-step explanation:
9(1-7x)
9-63x
Answer:
2, 0, 2, 3, 5
1, 2, 4, 0, 5
Step-by-step explanation:
(ax + b)(cx² + dx + e)
acx³ + adx² + aex + bcx² + bdx + be
2(2)x³ + 2(3)x² + 2(5)x + 0 + 0 + 0
4x³ + 6x² + 10x
a = 2
b = 0
c = 2
d = 3
e = 5
1(4)x³ + 1(0)x² + 1(10)x + 2(4)x² + 0 + 10
4x³ + 8x² + 10x + 10
a = 1
b = 2
c = 4
d = 0
e = 5
Ya, calculus and related rates, such fun!
everything is changing with respect to t
altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min
area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min
base=db/dt
alright
area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt
ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b
so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min
therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt
the base is decreasing at 1.6cm/min
Answer:
Uy
Step-by-step explanation:
