All categories

3 years ago

I know this is a lot but can anyone please help with 10-15?

Mathematics

2 answers:

aniked [119]3 years ago

8 0

Answer:

No we can't help

Step-by-step explanation:

You have to upload the entire sheet. We can't see the data we need to answer the questions

Natali5045456 [20]3 years ago

3 0

Answer:

what

Step-by-step explanation:

i din't understand this question

You might be interested in

Write a word phrase for each algebraic expression 5(x-2)

givi [52]

Five times the sum of x and negative two, or five times the difference of x and two

6 0

3 years ago

A trough is 8 feet long and has perpendicular cross section in the shape of an isosceles triangle (point down) with base 1 foot

murzikaleks [220]

Answer:

Dh/dt = 0.082 ft/min

Step-by-step explanation:

As a perpendicular cross section of the trough is in the shape of an isosceles triangle the trough has a circular cone shape wit base of 1 feet and height h = 2 feet.

The volume of a circular cone is:

V(c) = 1/3 * π*r²*h

Then differentiating on both sides of the equation we get:

DV(c)/dt = 1/3* π*r² * Dh/dt (1)

We know that DV(c) / dt is 1 ft³ / 5 min or 1/5 ft³/min

and we are were asked how fast is the water rising when the water is 1/2 foot deep. We need to know what is the value of r at that moment

By proportion we know

r/h ( at the top of the cone 0,5/ 2) is equal to r/0.5 when water is 1/2 foot deep

Then r/h = 0,5/2 = r/0.5

r = (0,5)*( 0.5) / 2 ⇒ r = 0,125 ft

Then in equation (1) we got

(1/5) / 1/3* π*r² = Dh/dt

Dh/dt = 1/ 5*0.01635

Dh/dt = 0.082 ft/min

4 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:

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