If someone used 1/4 of an ounce of nuts to make 1/12 of a recipe of cookies , how many ounces of nuts will she need to make one
1 answer:
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n[total]=25
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the number of possible outcome
=outcomes
Answers:
(a) f is increasing at
.
(b) f is decreasing at
.
(c) f is concave up at
(d) f is concave down at
Explanations:
(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that
So,
The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:
-->> x < -2
-->> -2 < x < 2
--->> x > 2
If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.
If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.
So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since
Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at.
(b) f is decreasing only when the derivative of f is negative. Since
Using the similar computation in (a),
Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.
Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)
(c) f is concave up if and only if the second derivative of f is positive. Note that
Since,
Therefore, f is concave up at.
(d) Note that f is concave down if and only if the second derivative of f is negative. Since,
Using the similar computation in (c),
Therefore, f is concave down at.
(a) f is increasing at
(b) f is decreasing at
(c) f is concave up at
(d) f is concave down at
Explanations:
(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that
So,
The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:
-->> x < -2
-->> -2 < x < 2
--->> x > 2
If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.
If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.
So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since
Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at
(b) f is decreasing only when the derivative of f is negative. Since
Using the similar computation in (a),
Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.
Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)
(c) f is concave up if and only if the second derivative of f is positive. Note that
Since,
Therefore, f is concave up at
(d) Note that f is concave down if and only if the second derivative of f is negative. Since,
Using the similar computation in (c),
Therefore, f is concave down at
Answer:
10. 105
11. 75
12. 75
13. 105
14. 75
15. 105
Step-by-step explanation:
Notice that 75? Anything that is adjacent (has a common vertex like 75º) to the vertex is a linear pair. here is an example below:
Since there is only THAT 75, you must find the supplementary angle (adds up to 180). and remember in order to get a supplementary angle it has to be a STRAIGHT LINE and HAS TO EQUAL 180. (4, 1 in your image)
180 - 75 = 105.
and also it's also a linear pair so it reflects too which makes things a lot easier!
So, if 4 is 75 and it reflects, then that means 2 is also 75. 1 is 105 and it reflects so 3 has to be 105. and the same for the rest of the numbers.
Hope this makes sense
~R3VO
