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alexdok [17]
2 years ago
13

A cylindrical container has a diameter of 5 inches and a height of 12 inches. The container will be wrapped and delivered as a g

ift. What measurement is closest to the amount of wrapping paper needed to cover the container without overlap?
Mathematics
1 answer:
Anna71 [15]2 years ago
5 0

Answer:

Step-by-step explanation:

1. Write your formula

2. Plug it in

3. Solve

these steps will help you out with all of your surface area problems.

Like this

Since it is a <u>diameter</u> you have to divide the diameter by 2. That will get you your radius. Then you write your formula for a cylinder. Your formula is...

V = Bh                            (Pie is also known as 3.14)

your big (B) is (pie times r squared) Then you add your height.

so your formula will look like this...

V = pie times r squared...

the full formula will be...

V = (3.14) times (2.5 squared) times the height which in this case is (12)

everything together will be...

V = (3.14) (2.5^2) (12)

all you have to do is multiply all of that and you will get your answer.

The answer is...

V = 236 inches ^2                        (Just a tip, the ^ sign means to the power of)

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Translate each into an algebraic expression
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Answer:

A) 5k

Step-by-step explanation:

anything near a number like that is usually times. Like 7g would be 7xg!

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3 years ago
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Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x &gt; 0)
vfiekz [6]

Answer:

Step-by-step explanation:

given a point (x_0,y_0) the equation of a line with slope m that passes through the  given point is

y-y_0 = m(x-x_0) or equivalently

y = mx+(y_0-mx_0).

Recall that a line of the form y=mx+b, the y intercept is b and the x intercept is \frac{-b}{m}.

So, in our case, the y intercept is (y_0-mx_0) and the x  intercept is \frac{mx_0-y_0}{m}.

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph (x_0,\frac{1}{x_0}). Which means that y_0=\frac{1}{x_0}

The slope of the tangent line is given by the derivative of the function evaluated at x_0. Using the properties of derivatives, we get

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Replacing the values in our previous findings we get that the y intercept is

(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}

The x intercept is

\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0

The triangle in consideration has height \frac{2}{x_0} and base 2x_0. So the area is

\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2

So regardless of the point we take on the graph, the area of the triangle is always 2.

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Gennadij [26K]

select two points on the line, and plug into this formula y2-y1 over x2-x1

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