Answer: 0, Zero
Step-by-step explanation: all horizontal lines are of the form "y = some number", and the equation "y = some number" always graphs as a horizontal line.
We have that
<span>(c-4)/(c-2)=(c-2)/(c+2) - 1/(2-c)
</span>- 1/(2-c)=-1/-(c-2)=1/(c-2)
(c-4)/(c-2)=(c-2)/(c+2)+ 1/(c-2)------- > (c-4)/(c-2)-1/(c-2)=(c-2)/(c+2)
(c-4-1)/(c-2)=(c-2)/(c+2)---------------- > (c-5)/(c-2)=(c-2)/(c+2)
(c-5)/(c-2)=(c-2)/(c+2)------------- > remember (before simplifying) for the solution that c can not be 2 or -2
(c-5)*(c+2)=(c-2)*(c-2)------------------ > c²+2c-5c-10=c²-4c+4
-3c-10=-4c+4----------------------------- > -3c+4c=4+10----------- > c=14
the solution is c=14
the domain of the function is (-∞,-2) U (-2,2) U (2,∞) or
<span>all real numbers except c=-2 and c=2</span>
Answer:
<h3>
<em><u>I </u></em><em><u>don't </u></em><em><u>know </u></em><em><u>what </u></em><em><u>it </u></em><em><u>is </u></em><em><u>that </u></em><em><u>you've </u></em><em><u>been </u></em><em><u>doing </u></em><em><u>cause </u></em><em><u>my </u></em><em><u>heart </u></em><em><u>stop </u></em><em><u>speeding </u></em><em><u>triple </u></em><em><u>times </u></em><em><u>when </u></em><em><u>the </u></em><em><u>causing </u></em><em><u>starts </u></em><em><u>with </u></em><em><u>you </u></em></h3>
Step-by-step explanation:
<h3>I get so weak under knees I lose all control and something takes over me and the days I've been so amazing it's not a face I want you to stay with me by my side my pride</h3>
Here is the expansion of <span>(x + y)n </span><span>for </span><span>n = 0, 1,…, 5</span><span> :</span>
<span>(x + y)0 = 1</span>
<span>(x + y)1 = x + y </span>
<span>(x + y)2 = x 2 +2xy + y 2 </span>
<span>(x + y)3 = x 3 +3x 2 y + 3xy 2 + y 3 </span>
<span>(x + y)4 = x 4 +4x 3 y + 6x 2 y 2 +4xy 3 + y 4 </span>
<span>(x + y)5 = x 5 +5x 4 y + 10x 3 y 2 +10x 2 y 3 +5xy 4 + y 5
</span>
hopefuly this helps you understand
