<span>Given that triangle
NLM is reflected over the line segment as shown, forming triangle ABC.
When a point is refrected across a line, the relative distance form the point to the line of refrection is preserved. That is the distance from the point to the line of refrection is equal to the distance of the image to the line of refrection.
Thus, from the figure, it can be seen the point B is of the same distance to the line of refrection as point M, so is point A to point L and point C to point N.
Thus, </span><span>ΔNLM is similar to </span><span><span>ΔCAB
Therefore, the</span> congruency statement that is correct is ΔNLM ≅ ΔCAB</span>
Let number of driveways , she shoveled on Sunday = x
And for 4 driveways , Kendra charges = 4*11=44
Therefore,
143 = 44 + 11x
And that's the model for the given situation .
Subtracting 44 from both sides,
143-44 = 11x
99 = 11x
Dividing both sides by 11
x = 9
Therefore she shoveled 9 driveways on Sunday .
There are
ways of selecting two of the six blocks at random. The probability that one of them contains an error is
So
has probability mass function
These are the only two cases since there is only one error known to exist in the code; any two blocks of code chosen at random must either contain the error or not.
The expected value of finding an error is then
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.
Answer:
765 J
Step-by-step explanation:
We are given;
Mass of bucket = 30 kg
Mass of rope = 0.3 kg/m
height of building= 30 meter
Now,
work done lifting the bucket (sand and rope) to the building = work done in lifting the rope + work done in lifting the sand
Or W = W1 + W2
Work done in lifting the rope is given as,
W1 = Force x displacement
W1 = (30,0)∫(0.2x .dx)
Integrating, we have;
W1 = [0.2x²/2] at boundary of 30 and 0
W1 = 0.1(30²)
W1 = 90 J
work done in lifting the sand is given as;
W2 = (30,0)∫(F .dx)
F = mx + c
Where, c = 30 - 15 = 15
m = (30 - 15)/(30 - 0)
m = 15/30 = 0.5
So,
F = 0.5x + 15
Thus,
W2 = (30,0)∫(0.5x + 15 .dx)
Integrating, we have;
W2 = (0.5x²/2) + 15x at boundary of 30 and 0
So,
W2 = (0.5 × 30²)/2) + 15(30)
W2 = 225 + 450
W2 = 675 J
Therefore,
work done lifting the bucket (sand and rope) to the top of the building,
W = 90 + 675
W = 765 J
