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max2010maxim [7]
3 years ago
12

How many different pairs of parallel edges are there on a rectangular solid?

Mathematics
1 answer:
Monica [59]3 years ago
4 0

Answer:

18

Step-by-step explanation:

A rectangular prism has four parallel edges along its length, four parallel edges along its width, and four parallel edges along its height.

We want to know how many different pairs of parallel edges there are.  Starting with the length, the number of unique pairs is:

₄C₂ = 6

The same is true for the width and height.  So the total number of different pairs of parallel edges is:

3 × 6 = 18

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Sphinxa [80]
If you meant 2x + 2x + 1 = 17: x = 4
Solve for x by simplifying both sides of the equation, then isolating the variable.<span> 
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If you meant <span>x2 + 2x + 1 = 17:</span> <span>x ≈ 3.1231056,−5.1231056x</span>
Solve the equation for x by finding a, b, and c of the quadratic then applying the quadratic formula. <span><span>x = −1 ± <span>√17</span></span>x</span>
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Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0
IrinaK [193]
We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: 36 y^{2} =( x^{2} -4)^3
We divide by 36 and take the root of both sides to obtain: y = \sqrt{ \frac{( x^{2} -4)^3}{36} }

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: y =  \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}

Let's leave that for the moment and look at the formula for arc length. The formula is L= \int\limits^c_d {ds} where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: ds= \sqrt{1+( \frac{dy}{dx})^2 } dx

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4). More formally, we can let u=x^{2} -4 and then consider the derivative of u^{3/2}du. Either way, we obtain,

\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}

This means that in our case:
ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx
ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx
ds=  \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx

Recall, the formula for arc length: L= \int\limits^c_d {ds}
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, [(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}


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Tarana mailed 20 greetings cards. 9 of them were mailed to USA. What percentage of the cards were mailed to USA?
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Answer:

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Step-by-step explanation:

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Answer:

<u>\fbox{ Correct Answer Option B \& D}</u>

Step-by-step explanation:

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I've solved all the equation for given both the points and concluded Option B & C are the correct equation for the line. The image of solution I'm attaching with this answer.

\small \sf Thanks  \: for \:  joining  \: brainly  \: community!

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