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jarptica [38.1K]
3 years ago
8

Someone, please help this is also for my final grades so please do it correctly:)

Mathematics
2 answers:
bija089 [108]3 years ago
5 0
In my opinion I think it’s 3/7
Trava [24]3 years ago
4 0

Answer:

well, There looking for a paint mixture of a pink. but we don't want to over power the white to mush, so we know that there has to be less red paint than white. My guess would be 3/7. I really hope this is correct and if it is i hope this helped.

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Physics students were modeling the height of a ball once it was dropped from the top of a 10 foot ladder. The
julsineya [31]

Answer:

<em>Interval notation: [0, 2.236]</em>

<em>Set Builder notation: </em>\{t\ |\ 0\le t\le 2.236\}

Step-by-step explanation:

Given that:

Equation of height of the ball dropped from a height of 10 foot, as:

h(t) = 10 - 2t^2

Where t is the time since the ball was dropped.

To find:

The domain of the function in Interval and set builder notation.

Solution:

<em>Domain of a function </em>is defined as the set of valid input values that can be given to the function for which the function is defined.

Here, input is time.

We can not have negative values for time.

Therefore, starting value for time will be <em>0 seconds</em>.

And the value of height can not be lesser than that of 0 ft.

0= 10 - 2t^2\\\Rightarrow 2t^2=10\\\Rightarrow t^2=5\\\Rightarrow t =2.236\ seconds

Maximum value for time can be 2.236 seconds.

Therefore the domain is:

<em>Interval notation: [0, 2.236]</em>

<em>Set Builder notation: </em>\{t\ |\ 0\le t\le 2.236\}

4 0
3 years ago
What is the domain and range?
ycow [4]

Answer:

Domain- the set of possible values of the independent variable or variables of a function.

Range- the set of values that a given function can take as its argument varies.

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:

<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
Consider the two similar cylinders. The heights of both cylinders are given, along with the diameter of the smaller cylinder. Wh
NeX [460]

Answer:

it is 18

Step-by-step explanation:

to find the scale factor divide 48 by 32 to get 1.5, and then multiply the diameter given which is 12, by 1.5 to get 18

3 0
3 years ago
!!!!HELP ME PLZZZZZZZZZZZZZ
Zolol [24]

Answer:

2) Yes

3) No

4) Yes

5) 2x + 8

6) 5(x - 8)

7) 3x + 21

Hope this helps :)

5 0
2 years ago
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