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Leona [35]
3 years ago
12

A farmer has a collapsable grain bag which can expand both vertically and horizontally as he needs it to. He wants to find a fun

ction to represent how much grain he can put in bag based on how much he has expanded it. When the bag is laying flat on the ground the area can be represented by the function f(x) = 3x2 + 1 and as the bag is raised up the height can be represented by the function g(x) = x+ 5. In terms of x, how much grain can the bag hold?
Mathematics
1 answer:
bija089 [108]3 years ago
6 0

Answer:

3x^{3}+15x^{2}+x+5

Step-by-step explanation:

We are told that the area of the bag can be represented by the function f(x)=3x^{2} +1 and as the bag is raised up its height can be represented by the function g(x)=x +5.

Since we know that we can find volume of cuboid by multiplying base area to its height. We are given area and height of bag as functions. Now we will find volume of the bag by multiplying these functions.  

\text{Volume of collapsible bag}=f(x)*g(x)  

\text{Volume of collapsible bag}=(3x^{2}+1)*(x +5)

After using distributive property we will get,

\text{Volume of collapsible bag}=3x^{2}(x+5)+1(x+5)

\text{Volume of collapsible bag}=3x^{3}+15x^{2}+x+5

Therefore, the collapsible bag can hold 3x^{3}+15x^{2}+x+5 grain.

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lys-0071 [83]

Answer:

$.75

Step-by-step explanation:

9 ÷ 12 = 0.75

.75 × 12 = 9

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Can somebody help me with this problem
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3 years ago
Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Complementary angles total 90°. Here, 90° angle BCE is divided by ray CF into two complementary angles. They necessarily have a total of 90°.

  ∠ECF and ∠BCF

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