Answer:
<u>First question answer:</u> The limit is 69
<u>Second question answer:</u> The limit is 5
Step-by-step explanation:
For the first limit, plug in 
in the expression 
, that's the answer for linear equations and limits.
So we have:
The answer is 69
For the second limit, if we do same thing as the first, we will get division by 0. Also indeterminate form, 0 divided by 0. Thus we would think that the limit does not exist. But if we do some algebra, we can easily simplify it and thus plug in the value 
into the simplified expression to get the correct answer. Shown below:
<em>Now putting 1 in 
gives us the limit:</em>
So the answer is 5
Answer:
x=-6
Step-by-step explanation:
Let x be the number
3x = x^2 -54
Subtract 3x
3x-3x = x^2 -3x-54
0 = x^2 -3x-54
Factor
What 2 numbers multiply to -54 and add to -3
-9*6 = -54
-9+6 = -3
0=(x-9) (x+6)
Using the zero product property
x-9=0 x+6=0
x=9 x=-6
The questions asks for the negative solution
x=-6
Answer:
Step-by-step explanation:
Move all terms to one side
4x^2 + 4x + 5 = 0
Plug into quadratic formula
x = (-4 <u>+</u> 
) / 2(4)
= (-4 <u>+</u> 
)/8
At this point, there are no real answers. The roots to this equation are complex.
x = (-4 <u>+</u> 8i)/8
= (-1 <u>+</u> 2i)/2
The standard form for the equation of a circle is :
<span><span><span> (x−h)^</span>2</span>+<span><span>(y−k)^</span>2</span>=<span>r2</span></span><span> ----------- EQ(1)
</span><span> where </span><span>handk</span><span> are the </span><span>x and y</span><span> coordinates of the center of the circle and </span>r<span> is the radius.
</span> The center of the circle is the midpoint of the diameter.
So the midpoint of the diameter with endpoints at (−10,1)and(−8,5) is :
((−10+(−8))/2,(1+5)/2)=(−9,3)
So the point (−9,3) is the center of the circle.
Now, use the distance formula to find the radius of the circle:
r^2=(−10−(−9))^2+(1−3)^2=1+4=5
⇒r=√5
Subtituting h=−9, k=3 and r=√5 into EQ(1) gives :
(x+9)^2+(y−3)^2=5
