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Veseljchak [2.6K]
4 years ago
9

Todd owns 3 surfboards and 2 wet suits. If he takes one surfboard and one wetsuit

Mathematics
1 answer:
zepelin [54]4 years ago
3 0

Answer:

The answer is  don't ask me to explain cuz i can't O.O

but there yea go :3

Step-by-step explanation:

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2. A farmer sells half of his yearling calf herd at the local stockyards. His average weight for each of the 30 head sold was 57
Degger [83]

Answer:

A

Step-by-step explanation:

Cause it right and pls dont remind me

7 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:

<u>Calculus</u>

Limits

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

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. 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 <em>x</em> = 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 <em>x</em> = 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 <em>x</em> = 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
What is the coefficient of the term 4/3yz
Yuri [45]
4/3

..........................
7 0
3 years ago
Complete the square to determine the maximum or minimum value of the function defined by the expression. x2 + 8x + 6
irga5000 [103]

Answer:

Minimum at (-4, -10)

Step-by-step explanation:

x² + 8x + 6

The coefficient of x² is positive, so the parabola opens upward, and the vertex is a minimum.

Subtract the constant from each side

x² + 8x = -6

Square half the coefficient of x

(8/2)² = 4² = 16

Add it to each side of the equation

x² + 8x + 16 = 10

Write the left-hand side as the square of a binomial

(x + 4)² = 10

Subtract 10 from each side of the equation

(x+ 4)² -10 = 0

This is the vertex form of the parabola:

(x - h)² + k = 0,

where (h, k) is the vertex.

h = -4 and k = -10, so the vertex is at (-4, -10).

The Figure below shows your parabola with a minimum at (-4, -10).

4 0
3 years ago
PLEASE ANSWER WILL MARK BRAINLIEST!!! <br><br><br> What is the slope of y = -4x – 3?
natali 33 [55]
I think -4
Because the form is y=ax + b
a is the slop so -4 is the answer.
4 0
3 years ago
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