What is 3,960 in a fraction

Standard issue passenger license plates in Ohio consist of three letters followed by four numbers. How many standard issue licen

dezoksy [38]

Answer:

The standard issue license plates that can be produced if there are no restrictions on the letters and numbers = 175760000

Step-by-step explanation:

If there are no restrictions, all numbers and letters are available to be used then. And with no restrictions, every number or letter can appear more than once.

There are 7 spaces available; 3 spaces for letters, 4 spaces for numbers

The different combination of letters and numbers then becomes,

26 × 26 × 26 × 10 × 10 × 10 × 10

This is because, all 26 letters (A to Z) can occupy the first space, the second space and the third space. And all 10 digits (0 to 9) can occupy the fourth space, the fifth space, the sixth space and the seventh space.

So, the standard issue license plates that can be produced if there are no restrictions on the letters and numbers = 26 × 26 × 26 × 10 × 10 × 10 × 10 = 175760000 different standard issue license plates.

8 0

2 years ago

I have 20 dollars, I want to split it between 3 people? How much would each person get if they got the same equal amount?

Ugo [173]

They would each get 6 dollars and 66 cents :) ♥♥

7 0

3 years ago

Read 2 more answers

The work shows how to use long division to find (x2 + 3x –9) ÷ (x – 2).

Maksim231197 [3]

2x+3x -9. You can simplify this to 5x-9.

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:

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

2 years ago

Write a system of equations and solve.

Scrat [10]

This looks a lot like a problem previously posted and already worked.
brainly.com/question/8829608

7 0

2 years ago