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2 years ago

How many digits are there in the repeating block for the decimal equivalent of 3 over 7?

Mathematics

1 answer:

Sindrei [870]2 years ago

7 0

Step-by-step explanation:

$\text{By long division, we know that}$

$\frac37=0.425871425871425871\ldots$

$\text{Observing the pattern, we know the repeating block is }\overline{425871}\text{. Therefore, there are 6 digits in the repeating block}$

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If a bird is on the ground, 7 ft away from a 10 ft tall tree, then what is the angle (to the

iren2701 [21]

Answer:

The angle is 55°

Step-by-step explanation:

To find the angle at which a bid is looking up on top of a tree that is 10feet tall if the bird is on the ground 7 feet away from the tree, we will follow the steps below:

we will use trigonometric ratio to find the angle

SOH CAH TOA

sinФ = opposite / hypotenuse

cosФ = adjacent / hypotenuse

tanФ = opposite /adjacent

from the diagram below, opposite = 10 feet and adjacent = 7 feet

the best trig.ratio to use is tan

tanФ = opposite /adjacent

tanФ = 10 /7

tan Ф =1.42857

to get the value of Ф, we will take the tan⁻¹ of both-side

tan⁻¹ tan Ф = tan⁻¹ 1.42857

Ф= tan⁻¹ 1.42857

Ф≈55°

Therefore, the angle is 55°

4 0

3 years ago

2 balls can complete a full rotation in 120 min. How many balls can complete a full rotation in 180 min

riadik2000 [5.3K]

Its 3, because each ball is 60 min each and 60 times 3 would get you 180

6 0

3 years ago

Zack delivers mail for the post office.everyday he walks a seven mile route.How many miles does Zack walk in one month if he wor

77julia77 [94]

If im right give me brainliest

Answer: 154

Step-by-step explanation: 22 represents days 7 represents miles he walks a day i think you to have to multiply. 22 times 7 is 154

3 0

2 years ago

Help please thank you

iogann1982 [59]

Equilateral

Isosceles

Scalene

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8 0

3 years ago

Find the derivative of following function.

Aleks04 [339]

Answer:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}$

General Formulas and Concepts:
Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
$\displaystyle (uv)' = u'v + uv'$

Derivative Rule [Quotient Rule]:
$\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

Step-by-step explanation:

*Note:

Since the answering box is way too small for this problem, there will be limited explanation.

Step 1: Define

Identify.

$\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}$

Step 2: Differentiate

We can differentiate this function with the use of the given derivative rules and properties.

Applying Product Rule:

$\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '$

Differentiating the first portion using Quotient Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Differentiating the second portion using Quotient Rule again:

$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule again:
$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Substitute in derivatives:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Simplify:

$\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}$

We can rewrite the differential by factoring and common mathematical properties to obtain our final answer:

∴ we have found our derivative.

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Learn more about derivatives: brainly.com/question/26836290

Learn more about calculus: brainly.com/question/23558817

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Topic: Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

8 0

2 years ago

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