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3 years ago

The table below represents the distance of a car from its destination as a function of time.

1 answer:

7 0

We are given the table that projects the distance of the car per hour. x is associated to time while y is associated to distance. The equation of the line after plotting in MS Excel is y = 900 - 50x. Hence the y-intercept is equal to 900 while the slope is equal to -50. The domain of the function that is x is all positive numbers.

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What is the additive identity of rational number

Burka [1]

Answer:

zero(0)

Step-by-step explanation:

The additive identity of a set of number is a number such that the its sum with any of the numbers in the set would give a result that is equal to the number in that set.

In other words, say for example the set of numbers is rational, the additive identity of rational numbers is 0. This is because, given any rational number say x, adding zero to the number x gives the same number x. i.e

x + 0 = x

If x is say 2, then we have;

2 + 0 = 2

Since adding zero to rational numbers gives has no effect on the numbers, then zero (0) is the additive identity of rational numbers.

3 0

3 years ago

PLEASE HELP ME ! (2,7); 3x-y =5 show all steps !

Zigmanuir [339]

Answer:

i dont think you've got the right (x,y) pair....

Step-by-step explanation:

3x - y = 5

3(2) - 7 = 5

6 - 7 = 5

those are not the correct coordinates for the equation

3 0

3 years ago

The product of the first 3 positive whole numbers

Kryger [21]

Answer:

If you mean by 1*2*3=6

Step-by-step explanation:

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3 years ago

Base your answer to the following question on The vertices of triangle ABC are A(1,2), B(3,8), and

Contact [7]

Answer:

Step-by-step explanation:

7 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