The function g(x) is defined as shown.
2 answers:
Answer:
3
Step-by-step explanation:
This is a piece-wise function. The domain [x-values] is broken down into parts.
<em>To evaluate this type of functions, we see in </em><em>WHICH PART</em><em> of the </em><em>DOMAIN</em><em> does the questions ask to look for.</em>
The question asks for , the value of the function at
<em>In which of the 3 parts does fall into?</em>
The domain condition, , in the last part tells us that it is GREATER THAN OR EQUAL TO 3. So
falls in this part of the function, which is
We can simply substitute 3 in of that part and get our desired answer. So,
The value of
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The Solution to the problem of the rectangle is given below
For Part A:
- equation model: (14 + w) * (w +8) = 132
- Length = 15
For Part B
- Perimeter = 46
Meaning of Perimeter
The perimeter of any shape can be defined as the total sum of the sides of the shape.
<h3>Analysis</h3>For Part A
equation model= (14 + w) * (w +8) = 135
14w + 112 + 8w + = 135
+ 22w - 23 = 0
solving the quadratic equation
w = -23 or 1
Because we are dealing with a physical quantity we will make use of the value 1
lenght of original rectangle = 10 + 1 = 11
Part B
Perimeter= 2(length) * 2 (breadth) = 2(11 + 4) * 2 (1 + 8)
Perimeter = 48
In conclusion, The Solution to the problem of the rectangle is given above.
Learn more about Perimeter: brainly.com/question/19819849
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Answer:
120
Step-by-step explanation:
We are given that the function for the number of students enrolled in a new course is .
It is asked to find the average increase in the number of students enrolled per hour between 2 to 4 hours.
We know that the average rate of change is given by,
,
where f(x)-f(a) is the change in the function as the input value (x-a) changes.
Now, the number of students enrolled at 4 = f(4) = = 255 and the number of students enrolled at 2 = f(2) =
= 15
So, the average increase =
=
= 120.
Hence, the average increase in the number of students enrolled is 120.
Answer:
The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.
Approximation at points (1.1,1.9) is 0.7
Step-by-step explanation:
Given:
Tangent plane to a surface z=5x+2y-10 as the function at point (1,2)
To find :
f(x,y) at (1,2)
partial derivatives of function w.r.t. (x and y) and value of that function at given points.
Solution:(refer the attachment also)
Now we know that
the equation of tangent plane at given points to the surface is given by,
f(x1,y1,z1) and z=f(x,y)
z-z1=Fx(x1,y1)*(x-x1)+Fy(x1,y1)*(y-y1)
here Fx(x1,y1) and Fy(x1,y1) are the partial derivatives of x and y.
now
taking partial derivative w.r.t. x we get
Fx(x1`,y1)=
=5.
Then w.r.t y we get
Fy(x1,y1)=
=2.
The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.
Using the Linearization or linear approximation we get
L(x,y)=f(x1,y1)+Fx(x,y)*(x-x1)+Fy(x,y)(y-y1)
=-1+5(x-1)+2(y-2)
=5x+2y-10
Approximation at F(1.1,1.9)
=5(1.1)+2(1.9)-10
=5.5+3.8-10
=0.7
Approximation at points (1.1,1.9) is 0.7
