All categories

2 years ago

X =0, y>=0 Maximum for P =3x +2y

Mathematics

1 answer:

givi [52]2 years ago

7 0

Answer:

day

Step-by-step explanation:

6th grade and the final grade for this semester and the

You might be interested in

<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

PLEASE HELP!! I NEED HELP WITH THIS PROBLEM!

valkas [14]

His basic salary is $ 2,500, his commission on sales is 3% (on sales only)
At the end of te month he got $3,040

Let x be the TOTAL sales performed during this month, so the money generated from sales (only) is (x).(3%), Hence the equation:

2500 + (x).(3%) = 3040
2500 + x(0.03) = 3040
x(0.03) = 3040-2500
x(0.03) = 540

and x = 540/0.03
x = $18,000

5 0

3 years ago

What is the range of the function y=4e^x

Harman [31]

Your answer is f(y) > 0
A is your answer

8 0

3 years ago

Read 2 more answers

PLEASE HELP!!! I WILL GIVE BRAINLIEST!!! Explain how to use estimation to help find the product of two decimals. Give an example

san4es73 [151]

You can use estimation to find the product of two decimals by rounding both the decimal’s so the nearest tenth or tens place (depending on how long it is) and then multiplying the decimals.

for example, if you had 4.6 and 8.9, you have to round the 4.6 and 8.9. you round the 4.6 up to 5 because the 6 bumps the 4 up to 5) and then round 8.9 to 9 (because the 9 bumps the 8 up to 8.) then, multiply 5 and 9 and you get 45!

6 0

2 years ago

Read 2 more answers

A trampoline has an area of 49pi square feet what's the diameter of the trampoline

Anna007 [38]

Area of a circle is π*r²
sub what we're given
49π= πr²
49π= π*r²
r²= 49π/π
r²=49
r=7
The diameter is twice the radius therefore the diameter is 14 units

7 0

3 years ago