Answer:
23.6°
Step-by-step explanation:
In this question we have to use some trigonometry to work out angle XVW. Since we are working out angles all of the trigonometric functions will have to be to the ⁻¹. The first thing we need to identify is will formula will we use out of the following:
Sin⁻¹ = Opposite ÷ Hypotenuse
Tan⁻¹ = Opposite ÷ Adjacent
Cos⁻¹ = Adjacent ÷ Hypotenuse
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10 = Hypotenuse
4 = Opposite
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We know which formula to use because the length won't be in the triangle for example we will be using the Sin triangle because we don't have an adjacent. If we don't have an adjacent then the other formula's won't work.
Now we substitute in the values to find the value of angle XVW
Sin⁻¹ = Opposite ÷ Hypotenuse
Sin⁻¹ = 4 ÷ 10
The value of angle XVW is 23.57817848
Answer:
720
Step-by-step explanation:
If every person has to choose a different character, the first person to choose a character has 6 options, the second has 5, the third has 4, the fourth has 3, and the last person has only two options. Therefore, the total number of ways you can do this if all 5 people dress up as a different character is:
![n=6*5*4*3*2\\n=720](https://tex.z-dn.net/?f=n%3D6%2A5%2A4%2A3%2A2%5C%5Cn%3D720)
There are 720 ways.
Answer:
option (3) is correct.
![e= \pm 3\sqrt{T}](https://tex.z-dn.net/?f=e%3D%20%5Cpm%203%5Csqrt%7BT%7D)
Step-by-step explanation:
Given ![T=\frac{e^2}{9}](https://tex.z-dn.net/?f=T%3D%5Cfrac%7Be%5E2%7D%7B9%7D)
We have to solve for e.
Consider the given statement,
![T=\frac{e^2}{9}](https://tex.z-dn.net/?f=T%3D%5Cfrac%7Be%5E2%7D%7B9%7D)
Cross multiply, we get,
![\Rightarrow 9T={e^2}](https://tex.z-dn.net/?f=%5CRightarrow%209T%3D%7Be%5E2%7D)
Taking square root both sides , we get,
![\Rightarrow \sqrt{9T}=e](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csqrt%7B9T%7D%3De)
We know square root of 9 is 3.
![\Rightarrow \pm 3\sqrt{T}=e](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cpm%203%5Csqrt%7BT%7D%3De)
Thus, option (3) is correct.
![e= \pm 3\sqrt{T}](https://tex.z-dn.net/?f=e%3D%20%5Cpm%203%5Csqrt%7BT%7D)
<h3>Refer to the diagram below</h3>
- Draw one smaller circle inside another larger circle. Make sure the circle's edges do not touch in any way. Based on this diagram, you can see that any tangent of the smaller circle cannot possibly intersect the larger circle at exactly one location (hence that inner circle tangent cannot be a tangent to the larger circle). So that's why there are no common tangents in this situation.
- Start with the drawing made in problem 1. Move the smaller circle so that it's now touching the larger circle at exactly one point. Make sure the smaller circle is completely inside the larger one. They both share a common point of tangency and therefore share a common single tangent line.
- Start with the drawing made for problem 2. Move the smaller circle so that it's partially outside the larger circle. This will allow for two different common tangents to form.
- Start with the drawing made for problem 3. Move the smaller circle so that it's completely outside the larger circle, but have the circles touch at exactly one point. This will allow for an internal common tangent plus two extra external common tangents.
- Pull the two circles completely apart. Make sure they don't touch at all. This will allow us to have four different common tangents. Two of those tangents are internal, while the others are external. An internal tangent cuts through the line that directly connects the centers of the circles.
Refer to the diagram below for examples of what I mean.
The maximum volume of the box is 40√(10/27) cu in.
Here we see that volume is to be maximized
The surface area of the box is 40 sq in
Since the top lid is open, the surface area will be
lb + 2lh + 2bh = 40
Now, the length is equal to the breadth.
Let them be x in
Hence,
x² + 2xh + 2xh = 40
or, 4xh = 40 - x²
or, h = 10/x - x/4
Let f(x) = volume of the box
= lbh
Hence,
f(x) = x²(10/x - x/4)
= 10x - x³/4
differentiating with respect to x and equating it to 0 gives us
f'(x) = 10 - 3x²/4 = 0
or, 3x²/4 = 10
or, x² = 40/3
Hence x will be equal to 2√(10/3)
Now to check whether this value of x will give us the max volume, we will find
f"(2√(10/3))
f"(x) = -3x/2
hence,
f"(2√(10/3)) = -3√(10/3)
Since the above value is negative, volume is maximum for x = 2√(10/3)
Hence volume
= 10 X 2√(10/3) - [2√(10/3)]³/4
= 2√(10/3) [10 - 10/3]
= 2√(10/3) X 20/3
= 40√(10/27) cu in
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