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2 years ago

Which number is the largest A. 0.37 B. 0.037 C. 0.004

Mathematics

2 answers:

DENIUS [597]2 years ago

8 0

The correct answer is A

son4ous [18]2 years ago

5 0

A . 0.37 is the largest it's closest to one

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Aidan bought a jacket on sale for $38.40. If the original price of the jacket was $64, what was the percent of the discount?

UkoKoshka [18]

Answer:

60%

Step-by-step explanation:

$64*.60= $38.40

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2 years ago

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Pls help me no explain need ❤️❤️❤️

RoseWind [281]

Answer:

$\displaystyle m=2$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Reading a Cartesian plane
Coordinates (x, y)
Slope Formula: $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Find points from graph.

Point (-5, 0)

Point (-9, -8)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute in points [Slope Formula]: $\displaystyle m=\frac{-8-0}{-9+5}$
[Fraction] Subtract/Add: $\displaystyle m=\frac{-8}{-4}$
[Fraction] Divide: $\displaystyle m=2$

3 0

2 years ago

Slope is 1/2 and (4,6) is on the line

AleksandrR [38]

Answer:

y-6=1/2(x-4)

Step-by-step explanation:

y-y1=m(x-x1)

y-6=1/2(x-4)

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2 years ago

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The set of ordered pairs represented by the graph below can be described as which of the following

Aneli [31]

ANSWER

a relation only

EXPLANATION

The given graph shown in the attachment represents only a relation and not a function.

The reason is that, a vertical line drawn across this graph will intersect this graph at more than one point.

Since the graph fails to pass the vertical line test, the ordered pair represented by this graph represents a relation only.

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2 years ago

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How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

2 years ago