Answer:
Step-by-step explanation:
are 1 and 2 answers becuas e if are 2
Answer:
k = 13The smallest zero or root is x = -10
===============================
note: you can write "x^2" to mean "x squared"
f(x) = x^2+3x-10
f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5
f(x+5) = (x^2+10x+25)+3(x+5)-10
f(x+5) = x^2+10x+25+3x+15-10
f(x+5) = x^2+13x+30
Compare this with x^2+kx+30 and we see that k = 13
x^2+13x+30 = 0
(x+10)(x+3) = 0
x+10 = 0 or x+3 = 0
x = -10 or x = -3
The smallest zero is x = -10 as its the left-most value on a number line.
Step-by-step explanation: Hope this helps kind of.
Please enclose the denominator in parentheses to reduce ambiquity.
<span>6x+24/x^2+7x+12
becomes
6(x+4) 6(x+4)
------------------ = ---------------- Cancel the factors (x+4)
(x^2+7x+12) (x+4)(x+3)
6
= -------------
x+3</span>
Answer: e. 1
Step-by-step explanation: a key problem in exponential smoothing is the choice of the values used for smoothing constants. It is easy to understand and quite easy to use, making it one of the most popular methods for forecasting. The forecast Ft+1 for the upcoming period is the estimate of average level Lt at the end of period t.
where α, the smoothing constant, is between 0 and 1. We can interpret the new forecast as the old forecast adjusted
by some fraction of the forecast error. The new estimate of level as a weighted average of
Dt (the most recent information on average level) and Ft (our previous estimate of that level).
Lt (and Ft+1 ) can be written in terms of all previous demand.
Thus, Ft+1 is a weighted average of all previous demand with the weight on Di given by α(1-α)t-i where t is the period
that just ended. As t increases the sum of these weights tends to 1.
Answer:
what exactly do you need help with?
Step-by-step explanation:
