Answer:
3/4
Step-by-step explanation:
Jessica shaded 6/8 of the circle. That fraction can be reduced by removing a factor of 2 from numerator and denominator. The reduced fraction is 3/4.
10 whole and 2 over 6 minus 7 whole and 3 over 6
2 answers:
Answer: 2 and 5/6
Explanation: To subtract mixed numbers, first subtract the fractions.
Notice here however that we have 2/6 - 3/6 which will gives us a negative fraction. Since this will cause us a lot of trouble, instead, let's rewrite the the first mixed number.
We can do this by thinking of 10 and 2/6 as 9 + 1 and 2/6 or as 9 + 8/6 by changing 1 and 1/6 into an improper fraction.
So 10 and 2/6 can be written as 9 and 8/6.
So we have 9 and 8/6 - 7 and 3/6.
Now subtract the fractions.
So we have 8/6 - 3/6 which is 5/6.
Then subtract the whole numbers. 9 - 7 is 2
2 and 5/6 is in lowest terms so 10 and 2/6 - 7 and 3/6 is 2 and 5/6.
Work is also attached in the image provided.
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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