Ruth is making lemonade. For 1 cup of water she uses 3 lemons, for 2 cups 6 lemons, etc. if this pattern continues, how many lem
2 answers:
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1. Ans:(A) 123
Given function:
The derivative would be:
=>
=>
=>
Now at x = 7:
=>
2. Ans:(B) 3
Given function:
The derivative would be:
=>
=>
=>
Now at x = 4:
(as constant)
=>Ans:
3
3. Ans:(D) -5
Given function:
The derivative would be:
or
=>
=>
Now at x = -1:
=>
=> Ans:
4. Ans:(C) 7 divided by 9
Given function:
The derivative would be:
or
=>
=>
Now at x = -3:
=>
=> Ans:
5. Ans:(C) -8
Given function:
Now if we apply limit:
=>
=> Ans:
6. Ans:(C) 9
Given function:
Now if we apply limit:
=>
=> Ans:
7. Ans:(D) doesn't exist.
Given function:
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.
Check:
If you apply the limit, answer would be infinity.
8. Ans:(A) Doesn't Exist.
Given function:
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.
9, 10.
Please attach the graphs. I shall amend the answer. :)
11. Ans:(A) Doesn't exist.
First We need to find out:
where,
If both sides are equal on applying limit then limit does exist.
Let check:
If x
9: answer would be 9+9 = 18
If x
9: answer would be 9-9 = 0
Since both are not equal, as
, hence limit doesn't exist.
12. Ans:(B) Limit doesn't exist.
Find out:
where,
If all of above three are equal upon applying limit, then limit exists.
When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8
ALL of the THREE must be equal. As they are not equal. 0
8; hence, limit doesn't exist.
13. Ans:(D) -∞; x = 9
Given function:
The derivative would be:
=>
=>
=>
Now at x = 7:
=>
2. Ans:(B) 3
Given function:
The derivative would be:
=>
=>
=>
Now at x = 4:
=>Ans:
3. Ans:(D) -5
Given function:
The derivative would be:
or
=>
=>
Now at x = -1:
=>
=> Ans:
4. Ans:(C) 7 divided by 9
Given function:
The derivative would be:
or
=>
=>
Now at x = -3:
=>
=> Ans:
5. Ans:(C) -8
Given function:
Now if we apply limit:
=>
=> Ans:
6. Ans:(C) 9
Given function:
Now if we apply limit:
=>
=> Ans:
7. Ans:(D) doesn't exist.
Given function:
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.
Check:
If you apply the limit, answer would be infinity.
8. Ans:(A) Doesn't Exist.
Given function:
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.
9, 10.
Please attach the graphs. I shall amend the answer. :)
11. Ans:(A) Doesn't exist.
First We need to find out:
If both sides are equal on applying limit then limit does exist.
Let check:
If x
If x
Since both are not equal, as
12. Ans:(B) Limit doesn't exist.
Find out:
If all of above three are equal upon applying limit, then limit exists.
When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8
ALL of the THREE must be equal. As they are not equal. 0
13. Ans:(D) -∞; x = 9
f(x) = 1/(x-9).
Table:
x f(x)=1/(x-9)
----------------------------------------
8.9 -10
8.99 -100
8.999 -1000
8.9999 -10000
9.0 -∞
Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)
14. Ans: -6
s(t) = -2 - 6t
Inst. velocity =
Therefore,
At t=2,
Inst. velocity = -6
15. Ans: +∞, x =7
f(x) = 1/(x-7)^2.
Table:
x f(x)= 1/(x-7)^2
--------------------------
6.9 +100
6.99 +10000
6.999 +1000000
6.9999 +100000000
7.0 +∞
Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)
-i
For the first circle, let's use the pythagorean theorem
now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.
so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.
when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.
now, let's check for second circle
well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.
let's check the third circle
this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.
so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.
when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.
now, let's check for second circle
well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.
let's check the third circle
this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.