Applying the segment addition theorem, the length of line segment VW is: 2 units.
<h3>What is the
Segment Addition Theorem?</h3>
The segment addition theorem states that the sum of the lengths of two segments that make up a larger line segment equals the measure of the larger line segment, if the point on the line segments are collinear.
UV = 8x
VW = x+1
UW = 10
UV + VW = UW (segment addition theorem)
Substitute the values
8x + (x + 1) = 10
Open bracket
8x + x + 1 = 10
Combine like terms
9x + 1 = 10
Subtract 1 from both sides
9x + 1 - 1 = 10 - 1
9x = 9
Divide both sides by 9
9x/9 = 9/9
x = 1
VW = 8x + (x + 1)
Plug in the value of x
VW = x + 1
VW = 1 + 1
VW = 2 units.
Therefore, applying the segment addition theorem, the length of line segment VW is: 2 units.
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Answer:
A
Step-by-step explanation:
v=πr2h
r=(3)²* 5
45π unit³
Answer:
9·x² - 36·x = 4·y² + 24·y + 36 in standard form is;
(x - 2)²/2² - (y + 3)²/3² = 1
Step-by-step explanation:
The standard form of a hyperbola is given as follows;
(x - h)²/a² - (y - k)²/b² = 1 or (y - k)²/b² - (x - h)²/a² = 1
The given equation is presented as follows;
9·x² - 36·x = 4·y² + 24·y + 36
By completing the square, we get;
(3·x - 6)·(3·x - 6) - 36 = (2·y + 6)·(2·y + 6)
(3·x - 6)² - 36 = (2·y + 6)²
(3·x - 6)² - (2·y + 6)² = 36
(3·x - 6)²/36 - (2·y + 6)²/36 = 36/36 = 1
(3·x - 6)²/6² - (2·y + 6)²/6² = 1
3²·(x - 2)²/6² - 2²·(y + 3)²/6² = 1
(x - 2)²/2² - (y + 3)²/3² = 1
The equation of the hyperbola is (x - 2)²/2² - (y + 3)²/3² = 1.
Answer:
36
Step-by-step explanation:
1-40% = 1-0.4=0.6 or 60%
60% not good at art.
60*60% = 60*6/10 = 6*6 = 36
Answer:
The population that gives the maximum sustainable yield is 45000 swordfishes.
The maximum sustainable yield is 202500 swordfishes.
Step-by-step explanation:
Let be , the maximum sustainable yield can be found by using first and second derivatives of the given function: (First and Second Derivative Tests)
First Derivative Test
Let equalize the resulting expression to zero and solve afterwards:
Second Derivative Test
This means that result on previous part leads to an absolute maximum.
The population that gives the maximum sustainable yield is 45000 swordfishes.
The maximum sustainable yield is:
The maximum sustainable yield is 202500 swordfishes.