PLEASE HELP!!! WILL MARK BRAINLIEST!!!

Between what two integers is 130? 10 and 11 11 and 12 12 and 13 13 and 14

Nostrana [21]

Answer:

13 and 14

Step-by-step explanation:

4 0

3 years ago

I bisects segment GH. If IH] = x2

Tasya [4]

52

Ajja iaha suna sus. Wis winw aha whw wh s

3 0

3 years ago

1. The ratio of boys to girls in Mr. Okafor's after-school club is the same as the ratio of boys to girls in Ms. Williams' after

Serjik [45]

27 is correct I hope you pass !

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

3 years ago

The Benjamin Franklin Parkway in Philadelphia, PA, runs diagonally from City Hall to the Philadelphia Museum of Art. In the scal

Ghella [55]

Answer:

6895ft

Step-by-step explanation:

Given data

The coordinates

City Hall is located at (11.4, 1.2) x1,y1= 11.4, 1.2

Museum of Art is located at (1.5, 10.8) x2,y2= 1.5, 10.8

Required:

the distance between the two points

Applying the Pythagorean theorem as

$d= \sqrt((x_2-x_1)^2+(y_2-y_1)^2)$ we can find the distance by substituting the given point coordinates

$d= \sqrt((1.5-11.4)^2+(10.8-1.2)^2)\\\\d= \sqrt((-9.9)^2+(9.6)^2)\\\\d= \sqrt(98.01+92.16)\\\\d= \sqrt190.17\\\\d=13.79$

in feet, we are told that a unit is 500 ft

then 13.79 will be 500*13.79

=6895.107ft

=6895ft

3 0

3 years ago