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zepelin [54]
3 years ago
7

If f(x)=3x+10x and g(x)=5x-3, find (f+g)(x)

Mathematics
1 answer:
galben [10]3 years ago
4 0

Answer:

8x+7

Step-by-step explanation:

(f+g)(x)=f(x)+g(x)=3x+10+5x-3=8x+7

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The volume of this cylinder is 20,403.72 cubic yards. What is the radius? Use ​ ≈ 3.14 and round your answer to the nearest hund
MakcuM [25]
The formula for the volume of a cylinder is V=3.14r^2(h).
So substituting the volume: 20,403.72=3.14r^2(h).

To solve for ‘r’, divide both sides by 3.14h
r^2=6,498/h
Then find square root of both sides
r= sqrt(6,498/h)
Since ‘h’ is not specified here, it cannot be solved further.
4 0
2 years ago
What is 12 divided by 3/4
Tpy6a [65]
12 divided by 3/4 =
16

<span>It is 16 because the reciprocal of 3/4 is 4/3 and 12 as a fraction is 12/1. So 12/1 x 4/3 is 48/3. And if you divide 48 by 3 you will get 16. If you don't get 16 then you did the wrong calculation.</span>

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8 0
4 years ago
Read 2 more answers
Max is observing the velocity of a runner at different times. After one hour, the velocity of the runner is 5 km/h. After two ho
Valentin [98]

You have velocity 1 of 5 km per hour after 1 hour and velocity 2 of 3 km per hour after 2 hours.


Part A)

Let Velocity = a +bt, with t being the time in hours.

For velocity 1 you have a +b = 5

For velocity 2, you have a + 2b =3

Now subtract velocity 1 from velocity 2:

a +2b - (a+b) = 3-5

Simplify to get b = -2

Now solve for a in the first equation:

a + b =5

Replace b with -2:

a - 2 = 5

a = 7

Now replace a and b in the original formula:

Velocity = 7 -2t


Part B)

Create a table with x and Y values to graph.

X axis would be the time and the velocity would be the Y axis

Your time would be 0 , 1, 2 3 and 4 hours.

Using the final equation from part A, replace t with the hours ans solve for the velocities, which become the y axis:

7 - 2(0) = 7-0 = 7

7 - 2(1) = 7-2 = 5

7 - 2(2) = 7-4 = 3

7 - 2(3) = 7-6 = 1

7-2(4) = 7-8 = -1


Now you have your x and y coordinates to plot on a graph:

(0,7) (1,5) (2,3) (3,1) and (4,-1)



7 0
4 years ago
Read 2 more answers
<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
Determine the selling price and amount of increase of a $270 bicycle with a 24% markup.
Vlad1618 [11]
Take the answer you get from that and add it to 270. That is your answer.
7 0
3 years ago
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