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FrozenT [24]
3 years ago
12

Nguyen is decorating the room for her sister’s baby shower party. She is wrapping Vietnamese conical hats in baby prints tissue

paper to decorate the tables and some shelves in the wall. Each hat has a diameter of 16 inches and a height of 9 inches. If she wants to wrap 21 Vietnamese hats, how much tissue paper will she need?
A 4,176 in2

B 8,975 in2

C 10,580 in2

D 36,267 in2
Mathematics
2 answers:
JulsSmile [24]3 years ago
6 0
So the best and correct answer is C because this is how you do it The surface area of the cone has formula

pi * radius * ( radius + sqrt ( height^2 + radius^2)) =
pi * radius^2 + pi * radius * sqrt( height^2 + radius^2)

However, this includes the base of the cone and the hats do NOT have
that. "The hats are shaped like a cone with NO BASE", says the 3rd sentence.

So the formula becomes pi * radius * sqrt( height^2 + radius^2)

diameter = 16
radius = 8
height = 9

The surface area of ONE of the hats is pi * 8 * sqrt( 9^2 + 8^2)
= pi * 8 * sqrt( 81+64)
= pi * 8 * sqrt(145)
302.63828053.....


21 of them is 21 * 302... = 6355.40689112841343251026086.....
Vedmedyk [2.9K]3 years ago
6 0

Answer:

The formula to calculate surface area of a conical hat in this case:

A = pi x radius x sqrt(radius^2 + height^2) + pi x radius^2

   = pi x (16/2) x sqrt((16/2)^2 + 9^2)) + pi x (16/2)^2

   = 503.7 (in2)

=> For 21 conical hats, Nguyen would need an amount of tissue paper:

A = 21 x 503.7 = 10577.7 (in2) = ~ 10580 (in2)

=> Option C is correct.

Hope this helps!

:)

Hi vọng là đáp án đúng!

:)

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The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t
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First Case:

\displaystyle p=\frac{5}{2}\text{ and } d=-2

Second Case:

\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

4p^2-10=(6p+2)+d

And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

d=4p^2-6p-12

And for the second equation:

2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

Therefore:

8p^2-12p-24=-2p+1

Simplifying yields:

8p^2-10p-25=0

Solve for <em>p</em>. We can factor:

8p^2+10p-20p-25=0

Factor:

2p(4p+5)-5(4p+5)=0

Grouping:

(2p-5)(4p+5)=0

Zero Product Property:

\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}

Then, we can use the second equation to solve for <em>d</em>. So:

2d_1=-2p_1+1

Substituting:

\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

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