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a new car is purchased for 207000 dollars. the value of the car depreciates at 11% per year. what will the value of the car be,

kicyunya [14]

Depreciation of 11% per year means that the price is multiplied by 89% (0.89) every year. Do this 11 times or use exponents

Final price:
207000*(0.89^11) = 57446.083

7 0

3 years ago

What is the median of the data set? 15, 19, 20, 22, 26, 30, 31

Paladinen [302]

Answer:

C. 22

Step-by-step explanation:

The median is the middle number of the set if the numbers are in order from least to greatest.

This set of numbers is in order, so the median is the middle number = 22

7 0

3 years ago

6x-y=-23 8x+3y=-9 Substition method (Show it step by step please!)

Tanya [424]

Answer:

solution is (- 3, 5 )

Step-by-step explanation:

given the 2 equations

6x - y = - 23 → (1)

8x + 3y = - 9 → (2)

rearrange (1) expressing y in terms of x

y = 6x + 23 → (3)

Substitute y = 6x + 23 into (2)

8x + 3(6x + 23 ) = - 9

8x + 18x + 69 = - 9

26x + 69 = - 9 ( subtract 69 from both sides )

26x = - 78 ( divide both sides by 26 )

x = - 3

substitute x = - 3 into (3)

y = (6 × - 3 ) + 23 = - 18 + 23 = 5

solution is (- 3, 5 )

5 0

3 years ago

WILL GIVE BRAINLIEST---

Sergeeva-Olga [200]

It's a rational number because 25 is perfect square, taking the square root gives the rational integers +5 or -5

3 0

3 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:

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