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2 years ago

Please help please thank you

Mathematics

1 answer:

AleksAgata [21]2 years ago

8 0

Honestly, I would go with:
C. Yes, because the mean is larger in the math book

Because the mean for the math book is 49.9 and the poetry book's mean is 41.5. And since 49.9 is a bigger number than 41.5, that is why i'm going with that one

But it's for sure one of the bottom 2

Hope this helps!! :-)

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Is 5:2 the same as 2:5

Neporo4naja [7]

Answer:

It is not the same

Step-by-step explanation:

Is 5/2 greater than 2/5? Is 5/2 bigger than 2/5? Is 5/2 larger than 2/5? These are all the same questions with one answer.

When comparing fractions such as 5/2 and 2/5, you could also convert the fractions (if necessary) so they have the same denominator and then compare which numerator is larger.

To get the answer, we first convert each fraction into decimal numbers. We do this by dividing the numerator by the denominator for each fraction as illustrated below: 5/2 = 2.5

2/5 = 0.4 Therefore, 5/2 is greater than 2/5 and the answer to the question "Is 5/2 greater than 2/5?" is yes.

6 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

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

Unit: Limits

6 0

3 years ago

The accompanying table shows the number of bacteria present in a certain culture over a 4 hour period, where x is the time, in h

Vlad [161]

Answer:

Step-by-step explanation:

I really hope you know how to use your calculator to find regression equations, because in this forum it is impossible to teach you how to do that. I ran the data through my calculator and got the exponential equation as

$y=1963.56(1.14)^x$ which means at 4 hours, we started with 1963 bacteria and it is growing at a rate of 114% each hour. Since we started out with an x value of 0, and that value represents the number of bacteria after 4 hours, then time 0 is 4 hours, which means that time 12 is 16 hours. We replace the x in the equation with a 12 and do the math:

$y=1963.56(1.14)^{12$ which gives us, in the end,

y = 9460

4 0

3 years ago

What is 18.75 rounded two decimals places

Serggg [28]

The answer would be 18.8

7 0

2 years ago

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katrin [286]

Answer:

x = 3, x = 0

Step-by-step explanation:

To solve these factored equations, just divide each part by zero and solve.

2x(x-3) = 0 → (x-3) = 0, and 2x = 0.

(x-3) = 0 → x - 3 + 3 = 0 + 3 → x = 3.

2x = 0 → 2x/2 = 0/2 → x = 0.

5 0

3 years ago