Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
Answer:

Step-by-step explanation:
7x + 24x = 25 x
31 x = 25 x
Subtracting 25x to both sides
31x - 25x = 0
6x = 0
Dividing both sides by 6
=> x = 0
Answer:
The missing reason in the proof is transitive property
Step-by-step explanation:
<u>Statement </u> <u>Reason </u>
1. x ∥ y w is a transversal 1. given
2. ∠2 ≅ ∠3 2. def. of vert. ∠s
3. ∠2 ≅ ∠6 3. def. of corr. ∠s
4. ∠3 ≅ ∠6 4. ??????????
From the statements 2 and 3
The previous proved statement to make use of the transitive property reason or proof
∴ 4. ∠3 ≅ ∠6 4. transitive property
Note: the transitive property states that: If a = b and b = c, then a = c.
Answer:
y=1/2+1
Step-by-step explanation:
1) Y2-Y1 over X2-X1
<u>9-5=4</u>
12-4=8
4/8 simplify the answer= 1/2
m=1/2
2) subtitute the simplification to one of the points.
y=mx+b
5=1(4)+b
5=4=b
1=b
y=1/2+1