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dalvyx [7]
3 years ago
11

PLEASE HELP!!! WILL MARK BRAINLIEST!!!

Mathematics
1 answer:
natka813 [3]3 years ago
7 0

Answer: the answer for all is 7.190

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Between what two integers is 130?<br> 10 and 11<br> 11 and 12<br> 12 and 13<br> 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
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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 !
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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
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
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