Answer:

<em><u>Linear function :</u></em>The characteristic property of linear functions is that when the input variable is changed, the change in the output is proportional to the change in the input. Linear functions are related to linear equations.
Step-by-step explanation:
We have
y=mx+c
for 1st
not satisfied.
for
2nd
not satisfied
<em><u>3rd</u></em>
<em><u>3rd satisfied</u></em>
4th
[note : substitute value of x to get value of y from table]
so
<u>t</u><u>h</u><u>i</u><u>r</u><u>d</u><u> </u><u>table represents a linear function.</u>
Answer:
Jade is faster than Ella by 2 minutes and 55 seconds.
Step-by-step explanation:
We are told that Ella and Jade both ran a marathon. Ella finished marathon in 4 hours 6 mins and 13 seconds. Jade finished marathon in 4 hours and 3 mins and 18 seconds.
We can write 4 hour 6 minutes 13 seconds as 4 hour 5 minutes 73 seconds. 1 minute has been borrowed to the seconds side, and seconds become 13+60 = 73 seconds.
We can see that Ella took more time than Jade. Let us find how much more time Ella took than Jade by subtracting time taken by Jade from time taken by Ella.
We can see that Jade took 2 minute and 55 seconds less than Ella , therefore, Jade was faster than Ella by 2 minutes 55 seconds.
Answer:

Step-by-step explanation:
Hi there!

Factor by grouping:

Let y=0. Apply the zero product property:


AND

I hope this helps!
Answer:
<em>The correct answer is: False</em>
Step-by-step explanation:
<u>If the sum of the opposite angles in a quadrilateral is 180°</u>, then a circle can be circumscribed about the quadrilateral.
Here, 
but, 
So, a circle can't be circumscribed about the given quadrilateral.