Think of the equation of a linear function:
Recall y = mx + b for vertical shifts, we just add or subtract from 'b' and that will move the line up or down accordingly.. However, for horizontal shifts, we will need to add or subtract from 'x'. Note that the slope or 'm' stays the same for each type of shift.
Now that we know how the shifts occur, we might consider a different form of the equation for a linear function: y = a(x - h) + k here the 'a' is just our slope, 'k' is our original y intercept, and 'h' will represent the amount of horizontal shift.
So to get the desired transformations of a horizontal shift to the left of 8 and a vertical shift of down 3 from our original function y = x, we can make the following changes: y = (x + 8) - 3 Now you might be confused with how we got the 'x + 8'.. Let's consider values of 'h'. For positive values of h, the result will be a shift to the right and for negative values of h the result will be a shift to the left. So since we want a shift to the left we need to use a '-8' and when we substitute that into our new form, y = (x - h) + k you can see the sign change.
Now we can simplify of course and get the final equation: y = x + 5 or in function form f(x) = x + 5
Answer:
see below
Step-by-step explanation:
Any line between two points on the circle is a chord.
Any angle with sides that are chords and with a vertex on the circle is an inscribed angle.
Any angle with sides that are radii and a vertex at the center of the circle is a central angle. Each central angle listed here should be considered a listing of two angles: the angle measured counterclockwise from the first radius and the angle measured clockwise from the first radius.
<h3>1.</h3>
chords: DE, EF
inscribed angles: DEF
central angles: DCF . . . . . note that C is always the vertex of a central angle
<h3>2.</h3>
chords: RS, RT, ST, SU
inscribed angles: SRT, RSU, RST, RTS, TSU
central angles: RCS, RCT, RCU, SCT, SCU, TCU
<h3>3.</h3>
chords: DF, DG, EF, EG
inscribed angles: FDG, FEG, DFE, DGE
central angles: none
<h3>4.</h3>
chords: AE
inscribed angles: none
central angles: ACB, ACD, ACE, BCD, BCE, DCE
Answer:
n+3
Step-by-step explanation:
(HoG)(x) = (2x)2 + 4 simply because HoG(x) is actually H(G(x)). So where ever there was an x in H(x) we substitute our value of G(x). Now the only thing left is to put in the 1. so out answer is (2(1))2 + 4 = 8Only in the 6th grade ; )
It’s probably A. (1,2) hope this helps..
