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lukranit [14]
3 years ago
7

What is 1/3, 1 1/4, 0.27, 1.35 in least to greatest

Mathematics
2 answers:
oee [108]3 years ago
5 0
<span>1/8,1/4,1/2,1,3/8,3/4,5/8,7/8</span>
Marina CMI [18]3 years ago
3 0

Answer:

It is D

0.27, 1/3, 1 1/4, 1.35

The key is

1. C

2. B

3. A

4. B

5. D

Step-by-step explanation:

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Please help me answer this question genius
ICE Princess25 [194]

(i) The Maclaurin's series is:

1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........

(ii) Integral of limit o to 1∫ e^sinx is 0

Maclaurin's expansion is as follows:

y(x) = y₀ ₊ x y₁(0) ₊ x²/2! y₂(0) ₊ x³/3! y₃(0) ₊ ........

y = e^sinx

y(0) = e^sin0

y(0) = 1

Differentiate y with respect to x

y₁(x) = e^sinx . cos x

y₁(0) = 1 [i.e y₁ = cosx.y]

y₁(0) = 1

Differentiate y₁ with respect to x

y₂(x) = cosc . y₁ ₋ sinx . y

∴ y₂(0) = 1 ₋ 0 = 1

Differentiate y₂ with respect to x

y₃ = cosx . y₂ ₋ (2 sinx ₊ cosx)y₁

y₃(0) = 1 ₋(0 ₊ 1)1

y₃(0) = 0

Differentiate y₃ with respect to x

y₄ = cosx . y₃ sinx . y₃ ₋ (2sinx ₊ cosx)y₂ ₋ (2cos x ₋ sinx) y₁

y₄(0) = 0 ₋ (3 . 0 ₊ 1)1 ₋ (2 ₋ o) 1

y₄(0) = -3

∴ From maclaurin's equation we get:

e^sinx = 1 ₊ x . 1 ₊ x²/2! . 1 ₊ x³/3! (0) ₊ x⁴/4! (₋3) ₊ ....

           =1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........

Hence we get the the series expansion as 1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........

(ii) limit 0 to 1∫e^sinx

integration formula ∫e^x dx = e^x + c

                    limit 0 to 1 ∫e^sinx = [e^sinx + c ]limit 0 to 1

                                                 = e^sin0 ₋ e^sin1

                                                 = 1 ₋ 1

                                                 = 0

Hence we get the answer as 0

Learn more about "Integration" here-

brainly.com/question/22008756

#SPJ10

8 0
1 year ago
Which place value is 1/10 the value of the ones place?
marishachu [46]

Answer:

the tenth place (0.1)

Step-by-step explanation:

4 0
3 years ago
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