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2 years ago

Find the 53rd term of the arithmetic sequence -12, -1, 10

Mathematics

2 answers:

RSB [31]2 years ago

5 0

440 is the answer

Tn=a+(n-1)d

Tn=53

a= -12

n=53

d= 11

-12+(53-1)11 = 440

fgiga [73]2 years ago

3 0

Answer:

Step-by-step explanation:

The sequence above is an arithmetic sequence

For an nth term in an arithmetic sequence

A(n) = a + ( n - 1)d

where a is the first term

n is the number of terms

d is the common difference

Since we're finding the 53rd term

a = - 12

n = 53

d = -1 - ( -12) = - 1 + 12 = 11 or 10--1 = 10 +1 =11

d = 11

So we have

A(53) = - 12 + (11)(53 - 1)

= - 12 + 11(52)

= -12 + 572

= 560

We have the final answer as

Hope this helps you

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Answer:

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Step-by-step explanation:

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Nataly_w [17]

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3 years ago

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I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

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$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

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$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

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Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

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• Compute the same integral using Green's theorem:

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$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

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3 years ago

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marysya [2.9K]

In this question, the first information that we get is that Kayla spent half of her weekly allowance in clothes. On cleaning the oven Kayla gets $4. Finally Kayla is left with $12.
Let us assume that the weekly allowance of Kayla = x
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x + (4 * 2) = 12 * 2
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x = 24 -8
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