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3 years ago

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Find the area of the region bounded by the curves y equals the inverse sine of x/4, y = 0, and x = 4 obtained by integrating wit

h respect to y. Your work must include the definite integral and the antiderivative.

1 answer:

olasank [31]3 years ago

3 0

Answer: 1/30

Step-by-step explanation:

∫[0,4] arcsin(x/4) dx = 2π-4

x = 4sin(u)

arcsin(x/4) = arcsin(sin(u)) = u

dx = 4cos(u) du

∫[0,4] 4u cos(u) du

∫[0,4] f(x) dx = ∫[0,π/2] g(u) du

v = ∫[1,e] π(R^2-r^2) dx

where R=2 and r=lnx+1

v = ∫[1,e] π(4-(lnx + 1)^2) dx

Using shells dy

v = ∫[0,1] 2πrh dy

where r = y+1 and h=x-1=e^y-1

v = ∫[0,1] 2π(y+1)(e^y-1) dy

v = ∫[0,1] (x-x^2)^2 dx = 1/30

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Step-by-step explanation:

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3 years ago

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irinina [24]

Angles C and D are supplementary, meaning they add up to 180 degrees. So, if we add 8u-48 to 5u+46, we get 13u-2. We set that equal to 180, so 13u-2=180. Add the two, so 13u=182. Divide the 13, so u=14. To double check, plug in 14 to both expressions. 8(14)-48 and 5(14)+46. 8(14)-48 is 64. 5(14)+46 is 116. If you add 64+116, you get 180, which proves your answer right! So u= 14

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3 years ago

The first, third and thirteenth terms of an arithmetic sequence are the first 3 terms of a geometric sequence. If the first term

Salsk061 [2.6K]

Answer:

The first three terms of the geometry sequence would be $1$ , $5$ , and $25$ .

The sum of the first seven terms of the geometric sequence would be $127$ .

Step-by-step explanation:

<h3>1.</h3>

Let $d$ denote the common difference of the arithmetic sequence.

Let $a_1$ denote the first term of the arithmetic sequence. The expression for the $n$ th term of this sequence (where $n\!$ is a positive whole number) would be $(a_1 + (n - 1)\, d)$ .

The question states that the first term of this arithmetic sequence is $a_1 = 1$ . Hence:

The third term of this arithmetic sequence would be $a_1 + (3 - 1)\, d = 1 + 2\, d$ .
The thirteenth term of would be $a_1 + (13 - 1)\, d = 1 + 12\, d$ .

The common ratio of a geometric sequence is ratio between consecutive terms of that sequence. Let $r$ denote the ratio of the geometric sequence in this question.

Ratio between the second term and the first term of the geometric sequence:

$\displaystyle r = \frac{1 + 2\, d}{1} = 1 + 2\, d$ .

Ratio between the third term and the second term of the geometric sequence:

$\displaystyle r = \frac{1 + 12\, d}{1 + 2\, d}$ .

Both $(1 + 2\, d)$ and $\left(\displaystyle \frac{1 + 12\, d}{1 + 2\, d}\right)$ are expressions for $r$ , the common ratio of this geometric sequence. Hence, equate these two expressions and solve for $d$ , the common difference of this arithmetic sequence.

$\displaystyle 1 + 2\, d = \frac{1 + 12\, d}{1 + 2\, d}$ .

$(1 + 2\, d)^{2} = 1 + 12\, d$ .

$d = 2$ .

Hence, the first term, the third term, and the thirteenth term of the arithmetic sequence would be $1$ , $(1 + (3 - 1) \times 2) = 5$ , and $(1 + (13 - 1) \times 2) = 25$ , respectively.

These three terms ( $1$ , $5$ , and $25$ , respectively) would correspond to the first three terms of the geometric sequence. Hence, the common ratio of this geometric sequence would be $r = 25 /5 = 5$ .

<h3>2.</h3>

Let $a_1$ and $r$ denote the first term and the common ratio of a geometric sequence. The sum of the first $n$ terms would be:

$\displaystyle \frac{a_1 \, \left(1 - r^{n}\right)}{1 - r}$ .

For the geometric sequence in this question, $a_1 = 1$ and $r = 25 / 5 = 5$ .

Hence, the sum of the first $n = 7$ terms of this geometric sequence would be:

$\begin{aligned} & \frac{a_1 \, \left(1 - r^{n}\right)}{1 - r}\\ &= \frac{1 \times \left(1 - 2^{7}\right)}{1 - 2} \\ &= \frac{(1 - 128)}{(-1)} = 127 \end{aligned}$ .

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2 years ago

Please help me find the answer

mixas84 [53]

136 would be the answer

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2 years ago

I need help pleaseeeeee

-BARSIC- [3]

Answer:

30

Step-by-step explanation:

total area of triangle is 180

so if 97+53= 150 180-150= 30

hope this helped

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3 years ago