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3 years ago

1. How many significant figures are there in 0.980 7?

Mathematics

1 answer:

beks73 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Sig figures are the total number of numbers to the right of the decimal point. For the answer, you count how many times it takes to get to the end of the number, in this case for .9807 you count 4.

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The temperature falls from 3°C to -4°C. What is the difference in these temperatures?

raketka [301]

Answer:

-1

Step-by-step explanation:

3- 4 = -1

3 0

4 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

Please help me with this fast, because i don't understand it

qaws [65]

Answer:

3/4

Step-by-step explanation:

f(0) - f(3) / 0 - 3

f(0) = -4

f(3) = 9/4 - 16/4 = -7/4

f(0) - f(3) = [-4 - (-7/4)]

f(0) - f(3) = [-16/4 + 7/4]

f(0) - f(3) = -9/4

-9/4 ÷ -3 = -9/4 · -1/3 = 3/4

3 0

3 years ago

There are 45 boys and 54 girls in your school’s senior class. What is the ratio of boys to girls

strojnjashka [21]

Answer:

5/6

Step-by-step explanation:

The ratio is:

45 / 54

Or, reduced:

5 / 6

7 0

4 years ago

Rewrite 8^1/3 using radical notation.

Aliun [14]

The answer is cube root 8

4 0

3 years ago