What is 54 to 15? Rationing math. What is the ratio of oil to vigar, oil is 54 vinegar is 15.

A train leaves Miami at 3:00pm. A second train leaves the same city in the same direction at 5:00pm. The second train travels 50

Alexxandr [17]

Let the speed of the first train be v mph. Then the speed of the second train is:
(v + 50) mph.
Distance traveled by the first train from 3pm to 7pm = 4v miles.
The second train travels 4v miles in 2 hours (from 5pm to 7pm). Therefore the speed of the second train = (4v)/2 = 2v.
However the speed of the second train is also given by (v + 50) mph.
v + 50 = 2v
v = 50.
The speed of the first train is 50 mph and the speed of the second train is 100 mph.

4 0

3 years ago

Which equation shows you how to find 16-9?

Marysya12 [62]

None shows the answer

6 0

3 years ago

You are purchasing a home that costs $240,000. You want to put 1/3 (or 33%) down. How much would you need for your down payment?

olchik [2.2K]

Answer:

$80,000

Step-by-step explanation:

240,000/3=80,000

80,000x1=80,000

5 0

3 years ago

Solve the equation x/3 -5 =y for x.

bagirrra123 [75]

A. X=3y + 15
I took the test last year

6 0

3 years ago

Find <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%20" id="TexFormula1" title=" \frac{dy}{dx} " alt=" \frac{d

nataly862011 [7]

Answer:

$\displaystyle y' = 2x + 3\sqrt{x} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

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:

Step 1: Define

Identify

$\displaystyle y = (x + \sqrt{x})^2$

Step 2: Differentiate

Chain Rule: $\displaystyle y' = 2(x + \sqrt{x})^{2 - 1} \cdot \frac{d}{dx}[x + \sqrt{x}]$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = 2(x + x^{\frac{1}{2}})^{2 - 1} \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]$
Simplify: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]$
Basic Power Rule: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 \cdot x^{1 - 1} + \frac{1}{2}x^{\frac{1}{2} - 1})$
Simplify: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2}x^{-\frac{1}{2}})$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2x^{\frac{1}{2}}})$
Multiply: $\displaystyle y' = 2[(x + x^{\frac{1}{2}}) + \frac{x + x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}]$
[Brackets] Add: $\displaystyle y' = 2(\frac{2x + 3x^{\frac{1}{2}} + 1}{2})$
Multiply: $\displaystyle y' = 2x + 3x^{\frac{1}{2}} + 1$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = 2x + 3\sqrt{x} + 1$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

4 0

3 years ago