The point P has coordinates (x,y) = (-2,6) so x = -2 and y = 6
Replace x and y with those values into the rule given
So,
(x,y) ---> (x-2, y-16)
turns into
(-2,6) ---> (-2-2, 6-16) = (-4,-10)
P = (-2,6)
P ' = (-4,-10)
The answer is -10 because your teacher just wants the y coordinate of point P'
<span>Let
x = the 10s digit
y = the units
N1----------> original number
</span>N2----------> new number
<span>N2=N1+36--------- >N2-N1=36
y+x=8------------- > equation (I)
then
10x + y = N1
10y+x=N2
(10y+x)-(10x+y)=36- -----> 9y-9x=36------- > y-x=4----------- > equation (II)
resolving (I) and (II)
</span> y+x=8
y- x=4<span>
-----------------
2y+0x=12--------------- >y=6
then
x=8-y=8-6=2----------> x=2
N1=26 and N2=62
the two digit number is 26</span>
-8y+10=4
-8y=-10+4
-8y=-6
Y=0.75
Option C: 
The last two terms given of the standard polynomial are:
This shows 2 patterns to us:
r 's degree is increasing per term in the given polynomial.
s 's degree is decreasing per term in the given polynomial.
The sum of powers of variates r and s is 2+4 = 3+3 = 6.
And since standard polynomial, the same pattern must be present in all terms of the series.
All options except C are not valid.
Option C has term for which r's degree plus s's degree is 1 + 5 = 6.
And r's degree 1 < s's degree 5.
Thus its a valid term to be putted as first term in standard form of polynomial in consideration.
Thus Option C: is the needed expression.
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