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3 years ago

6

If start fraction 1 over 3 end fraction is equivalent to 33start fraction 1 over 3 end fraction%, what percent is equivalent to

two-thirds? A. 33two-thirds% B. 150% C. 66two-thirds% D. 65%

1 answer:

kotegsom [21]3 years ago

5 0

Answer:

The answer is C 66%

Step-by-step explanation:

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10. Melissa has $20 to buy bagels and juice for her class. box of bagels bottle of juice 56.13, including tax $2.08, including t

adelina 88 [10]

The inequality for the maximum number of bottles of juice she can buy

6.13+ 2.08b <20.00

Option B

Given

Melissa has $20 to buy bagels and juice for her class

Box of bagels : 6.13 including tax

Bottle of juice: 2.08 including tax

She will buy only one box of bagels that is for 6.13

Let 'b' be the number of bottles of juice she can buy

Total cost should be less than 20 dollars

6.13(one bagels) +2.08(bottles of juices)<20

So the inequality becomes

$6.13+2.08b$

Learn more : brainly.com/question/1836165

4 0

2 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

<u>Taylor series</u> expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$

$\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4$

$\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4$

$\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4$

Substituting the values in the series expansion gives:

$e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...$

Factoring out e⁴:

$e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]$

<u>Taylor Series summation notation</u>:

$\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n$

Therefore:

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

7 0

1 year ago

What’s the volume of a cylinder with a base of 25 and height of 6

Anastasy [175]

Approximately 2945.24 units

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2 years ago

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Susan bought a 2 kg bag of grapes on the way home she ate 125 g of the grapes how many grams of grapes that she have left

amid [387]

There are 1,000 grams in a kilogram and she has two kilograms, so there are 2,000 grams. She eats 125 of them so you subtract 125 from 2,000 and you get 1875. She has 1875 grams left of grapes.

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3 years ago

Mrs. Barlow's math class took a test yesterday. Out of 70 students, 70% of them passed. How many of Mrs. Barlow's students passe

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The answer is A . 49

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3 years ago