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4 years ago

What is the slope for the function y=2x^2+2 at the point x= 3?

1 answer:

3 0

1) We have to calculate the first derivative:
y=2x²+2
y´=4x

2) the slope at the point x=3; will be f´(3);
f´(3)=4(3)=12

answer: the slope will be 12.

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I need help solving this problem |5x - 1| < 1

VashaNatasha [74]

Imagine what the graph would look like. Does it open upwards or downwards?

5x - 1 has a positive slope of 5 and it is reflected horizontally. The graph would be shaped like a V. A limited segment, or none, of the graph is below a certain value of y, and the rest outside it is above.

Hence, the solution must be a < x < b.

|5x - 1| < 1

5x - 1 < 1

x < 2/5

|5x - 1| < 1

5x - 1 > -1

x > 0

0 < x < 2/5

4 0

3 years ago

What value of x is in the solution set of 8x - 6 > 12 + 2x?

Alexxx [7]

Answer:

x>3

Step-by-step explanation:

7 0

3 years ago

A restaurant offers a catering service which costs $25.00 per person with a $119.50 service charge. For parties of 50 or more pe

Molodets [167]

Answer:

T1(n) = 25n +119.50 where n belongs to $1 \leq n$ .

T2(n) = 20n + 59 where n belongs to $50\leq n$ .

Step-by-step explanation:

We are given two situations first is for normal days and the other is for days when 50 or more people are going to this restaurant for the party.

Let the number of person going for this restaurant be "n"

For normal day, we are given

catering service = $25 per person

service charge = $119.50

Let the total cost of the restaurant be T

Cost for this case will be T1

T1(n) = 25n +119.50 where n belongs to $1 \leq n$

similarly for the case when 50 or more than that goes to this restaurant for party.

cost after discount

Catering service = $20.00 per person

service charge = $59.00

now the cost of this case will be T2

T2(n) = 20n + 59 where n belongs to $50\leq n$ .

Therefore the two piecewise linear functions represents the total cost of the restaurant for serving n people.

7 0

3 years ago

What is d=40 + 428t − 16t2

Veseljchak [2.6K]

Answer:

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3 0

2 years ago

Find the dimensions of the shaded region so that its area is maximized.<br> x=<br> y=

musickatia [10]

1. Find the equation of the line AB. For reference, the answer is y=(-2/3)x+2.
2. Derive a formula for the area of the shaded rectange. It is A=xy (where x is the length and y is the height).
3. Replace "y" in A=xy with the formula for y: y= (-2/3)x+2:

A=x[(-2/3)x+2] This is a formula for Area A in terms of x only.
4. Since we want to maximize the shaded area, we take the derivative with respect to x of A=x[(-2/3)x+2] , or, equivalently, A=(-2/3)x^2 + 2x.
This results in (dA/dx) = (-4/3)x + 2.
5. Set this result = to 0 and solve for the critical value:

(dA/dx) = (-4/3)x + 2=0, or (4/3)x=2 This results in x=(3/4)(2)=3/2

6. Verify that this critical value x=3/2 does indeed maximize the area function.
7. Determine the area of the shaded rectangle for x=3/2, using the previously-derived formula A=(-2/3)x^2 + 2x.

The result is the max. area of the shaded rectangle.

3 0

3 years ago