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2 years ago

Of the 400 words that brianna wrote for her book report

Mathematics

1 answer:

Black_prince [1.1K]2 years ago

8 0

The question here is a little unclear. Would love to help but please clarify :)

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(-4)(5)+(-3)(-2) whats is it???

Vlada [557]

Answer:

-14

Step-by-step explanation:

(-4)(5)+(-3)(-2)

=-20+6

=-14

5 0

3 years ago

Read 2 more answers

1-5x + 3y 45 2x+3y = 24

Norma-Jean [14]

Answer:

The answer is 1-5x+135y

Step-by-step explanation: multiply the numbers 3 and 45 which give you 135

3 0

3 years ago

If the ratio of the sides of two squares is 3:1, what is the ratio of their perimeters?

Anvisha [2.4K]

Answer:

Ratio of the perimeters =3:1

Step-by-step explanation:

We have given that : Ratio of the sides of two squares is 3:1

To find : Ratio of their perimeters

Solution : Let the length of the sides are 3:1 = 3x : x

Formula of perimeter of square = 4(side)

Using the formula ,

Perimeter of 1 square = 4×3x= 12x

Perimeter of 2 square = 4×x= 4x

Ratio of the perimeter of 1 square and 2 square = 12x : 4x

= 3 : 1

8 0

3 years ago

<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

2 years ago

Question 3 (Multiple choice) If 0

nikitadnepr [17]

Answer:

what on earth are you trying to ask

Step-by-step explanation:

thank you for reading this
and how do you time to read this
okay it's time to go baiiiiiiiiii

7 0

2 years ago