Answer:
a
Step-by-step explanation:
10 x the difference ofa number"n" - 6
First one = y6
second one = y -3
third one = y0
fourth one = y8
Claim: The difference between two rational numbers always is a rational number
Proof: You have a/b - c/d with a,b,c,d being integers and b,d not equal to 0.
Then:
a/b - c/d ----> ad/bd - bc/bd -----> (ad - bc)/bd
Since ad, bc, and bd are integers since integers are closed under the operation of multiplication and ad-bc is an integer since integers are closed under the operation of subtraction, then (ad-bc)/bd is a rational number since it is in the form of 1 integer divided by another and the denominator is not eqaul to 0 since b and d were not equal to 0. Thus a/b - c/d is a rational number.
Answer:
Go to the left nine Units and go up two units
Step-by-step explanation:
(A) <em>f(x)</em> = 7 is constant, so <em>f(x</em> + <em>h)</em> = 7, too, which makes <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 0. So <em>f'(x)</em> = 0.
(B) <em>f(x)</em> = 5<em>x</em> + 1 ==> <em>f(x</em> + <em>h)</em> = 5 (<em>x</em> + <em>h</em>) + 1 = 5<em>x</em> + 5<em>h</em> + 1
==> <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 5<em>h</em>
Then

(C) <em>f(x)</em> = <em>x</em> ² + 3 ==> <em>f(x</em> + <em>h)</em> = (<em>x</em> + <em>h</em>)² + 3 = <em>x</em> ² + 2<em>xh</em> + <em>h</em> ² + 3
==> <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 2<em>xh</em> + <em>h</em> ²

(D) <em>f(x)</em> = <em>x</em> ² +<em> </em>4<em>x</em> - 1 ==> <em>f(x</em> + <em>h)</em> = (<em>x</em> + <em>h</em>)² + 4 (<em>x</em> + <em>h</em>) - 1 = <em>x</em> ² + 2<em>xh</em> + <em>h</em> ² + 4<em>x</em> + 4<em>h</em> - 1
==> <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 2<em>xh</em> + <em>h</em> ² + 4<em>h</em>
