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Elza [17]
3 years ago
9

Tìm các giá trị của tham số m để hàm số y = (m-1)x^3 - 3(m-1)x^2 + 3(2m-3)x + m nghịch biến trên R

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
6 0

Answer:

3

Step-by-step explanation:

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Mjhs is refinishing a basketball court and decided to paint everything shown below blue. How many square feet will be painted bl
SashulF [63]

Answer:

<h2>Mjhs needs 189.25 square feet of blue paint, approximately.</h2>

Step-by-step explanation:

As you can observe in the image attached, the basketball court is a composite figure formed by a rectangle and half of a circle.

The rectangle has dimensions of 10 feet by 15 feet. The circle has a diameter of 10 feet.

The expression of this composite area would be

A= w \times l + \frac{\pi (\frac{l}{2} )^{2} }{2}

Where w = 15 \ ft and l = 10 \ ft. Replacing these values, we have

A= 15 \times 10 + \frac{3.14 (\frac{10}{2} )^{2} }{2}\\ A=150 + \frac{3.14 \times 25}{2}=150+39.25 \\ A \approx 189.25 \ ft^{2}

Therefore, Mjhs needs 189.25 square feet of blue paint, approximately.

8 0
3 years ago
In the number 7,777.777, how does the 7 in the hundreds place compare to the 7 in the place to its left?
Ivahew [28]
B is your answer!
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6 0
3 years ago
100 <br> 10 100 100 10 100 )))))))))
VashaNatasha [74]
Thanks for the points.
3 0
3 years ago
Read 2 more answers
At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
An airplane is loaded with cargo and ready to take off. Before it departs, an unknown number
Leto [7]

Answer:

b ≤ 50

Step-by-step explanation:

Since no more than 400 pounds of b number of 8 pound boxes can be taken off to ensure that the plane is in balance, this can be represented by the inequality:

8b ≤ 400

b ≤ 400/8

b ≤ 50

This means that a maximum number of 50 8 pounds boxes can be taken off the plane for a 600 pound crate to be put on ensuring that the plane is in balance.

8 0
3 years ago
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