Answer:
A (8,0)
B (8,8)
C (6,8)
D (6,0)
Step-by-step explanation:
Im pretty sure your can just flip the negative signs for "x" cause your flipping it across the y axis which changes the sign.
Answer:
=
−
1
1
Step-by-step explanation:
Answer:
in steps
Step-by-step explanation:
DE // BC
m∠ADE = m∠ABC and m∠AED = m∠ACB
∴ ΔADE similar to ΔABC
AB/AD = AC/AE
(AD + DB) / AD = (AE + EC) / AE
AD/AD + DB/AD = AE/AE + EC/AE
1 + DB/AD = 1 + EC/AE
DB/AD = EC/AE (AD/DB = AE/EC)
Answer:
b. y = 3(x+2)^2 + 5
Step-by-step explanation:
The vertex form of a parabola is
y = a(x - h)^2 + k where (h, k) is the vertex.
Here h = -2 and k = 5 so the following could be the required equation\;
y = 3(x+2)^2 + 5
Well this is simple a calculator type problem...but if you are curious as the the algorithm used by simple calculators and such...
They use a Newtonian approximation until it surpasses the precision level of the calculator or computer program..
A newtonian approximation is an interative process that gets closer and closer to the actual answer to any mathematical problem...it is of the form:
x-(f(x)/(df/dx))
In a square root problem you wish to know:
x=√n where x is the root and n is the number
x^2=n
x^2-n=0
So f(x)=x^2-n and df/dx=2x so using the definition of the newton approximation you have:
x-((x^2-n)/(2x)) which simplifies further to:
(2x^2-x^2+n)/(2x)
(x^2+n)/(2x), where you can choose any starting value of x that you desire (though convergence to an exact (if possible) solution will be swifter the closer xi is to the actual value x)
In this case the number, n=95.54, so a decent starting value for x would be 10.
Using this initial x in (x^2+95.54)/(2x) will result in the following iterative sequence of x.
10, 9.777, 9.774457, 9.7744565, 9.7744565066299210578124802523397
The calculator result for my calc is: 9.7744565066299210578124802523381
So you see how accurate the newton method is in just a few iterations. :P
