Answer:
Poor Alex, I know how they feel, I got a dog myself.
The amount of time between
11:41 pm to 12:09 am is
19 minutes (to 12 am, 60 minutes to next hour) + 9
which is 28 minutes for the first dog barking session
the second time is 35 minutes,
using same method as before.
adding the two separte barking sessions we get
35+28= 63 minutes
which is equivelant to 1 hour and 3 minutes.
In conclusion the dog barks for 1 hour and 3 minutes in total.
Answer:
Step-by-step explanation:
Real life = 8m = 8000mm
Drawing = 4mm
we can take the ratio of the two values (but be sure that they both have the same units, this is why it is necessary to convert 8m to 8000mm OR you can convert 4mm to its value in meters)
we can also write this as:
so the drawing's scale factor is:
This means that if length of the living room in the drawing is multiplied by 2000 it will be of the same length as the living room in real life.
The volume generated by rotating the region in the first quadrant bounded by y = ex and the x-axis from x = 0 to x = ln(3) about the y-axis is (πln(3)3)/3.
To find the volume generated by rotating the region in the first quadrant bounded by y = ex and the x-axis from x = 0 to x = ln(3) about the y-axis, we can use the disk method. The disk method involves slicing the region into thin disks and adding up their volumes.
The volume of each disk is πr2h, where r is the radius of the disk and h is the thickness of the disk. In this case, the radius of each disk is x and the thickness is dx.
So, the volume of the region is:
V = ∫0ln(3)πx2dx
We can use the power rule for integration to solve this integral:
V = π∫0ln(3)x2dx = π[(x3)/3]0ln(3) = π[(ln(3)3)/3 - (03)/3] = (πln(3)3)/3
Therefore, the volume generated by rotating the region in the first quadrant bounded by y = ex and the x-axis from x = 0 to x = ln(3) about the y-axis is (πln(3)3)/3. This is the exact form of the volume.
To know more about disk method refer here:
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The value of h that satisfies the equation |h-8|=|h+10| is -1.
To solve the equation |h - 8| = |h + 10|, we need to consider two cases based on the absolute values.
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B. x = 6 + 10; x = 16
C. x + 6 = 16; x = 10
D. x + 6 = 10; x = 4
Answer:
Step-by-step explanation:
to balance the beam, the weight on the left side must be equal to the right side.
x + 6 = 10 let x = box
x = 10 - 6
x = 4
therefore, the answer is D. x + 6 = 10; x = 4
x + 6 = 10; x = 4 is the linear equation and the solution that represents the model, where circles and a square are shown evenly balanced on a balance beam.
1. **Modeling the Balance**: In this problem, you are given a scenario where circles and a square are evenly balanced on a balance beam. To represent this balance mathematically, you need to ensure that the total weight (or the "value" of the shapes) on the left side of the balance is equal to the total weight on the right side.
2. **UnknownWeight**: You are asked to find the value of the square, represented by 'x.' This 'x' represents the weight or value of the square on one side of the balance.
3. **Equation Setup**: To set up the equation, you note that on the left side of the balance, you have 'x' (the square) plus 6 (the circles). On the right side, you have 10 (implying there's something with a weight or value of 10 units).
So, you set up the equation as:
x + 6 = 10
This equation says that the weight of the square plus the weight of the circles on one side equals the weight of whatever is on the other side.
4. **Solving for x**: To find the value of 'x' (the weight of the square), you isolate 'x' on one side of the equation. To do that, you subtract 6 from both sides of the equation:
x + 6 - 6 = 10 - 6
This simplifies to:
x = 4
So, 'x' represents a square with a weight or value of 4 units. This means that the square's weight on one side of the balance is balanced by 6 units of weight from the circles on the other side, resulting in a total of 10 units on both sides, ensuring the balance.
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The model rocket is in the air for 0 seconds according to the given equation, which implies an immediate impact upon launch.
In this mathematical scenario, the model rocket's height p(x) over time x (elapsed seconds) is given by the quadratic equation p(x) = 16x^2 + 32x. The total amount of time the rocket is in the air will be the point when the rocket returns to ground level. This occurs when p(x) = 0, which represents the rocket's height being zero feet above the ground.
We can find out when this occurs by solving the quadratic equation for x. We can rearrange the quadratic equation to 16x^2 + 32x = 0. Factoring out 16x gives us 16x(x + 2) = 0. Solving for x will give two potential solutions: x = 0 (the initial launch point) and x = -2. However, since time cannot be negative in this context, we discard the -2 and our answer is x=0 s, the total time the model rocket will be in the air after being launched is 0 seconds.
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