The map uses a scale of 1 inch miles. You are making a map with a scale of . How far apart are cities A and B on your map?

1 answer:

4 0

Answer:

You need to show a map and put the entire directions.

Step-by-step explanation:

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morpeh [17]

Answer:

w = 0, x = 3, y = 2, z = 1

Step-by-step explanation:

You can use each equation to write an expression that will substitute into the next equation.

y = 8 -2x . . . . from the first equation

8 -2x + 3z = 5 . . . . . substituted into the second equation

z = (-3 +2x)/3 . . . . . solved for z

(-1 +2/3x) +2w = 1 . . . . . substituted into the third equation

w = (2 -2/3x)/2 . . . . . solved for w

5(1 -1/3x) +3x = 9 . . . . . substituted into the last equation

4/3x = 4 . . . . . simplified, 5 subtracted

x = 3 . . . . . . . . multiply by 3/4

w = 0 . . . . . . . .value of x substituted into equation for w

z = 1 . . . . . . . . value of x substituted into equation for z

y = 2 . . . . . . . .value of x substituted into equation for y

The solution is (w, x, y, z) = (0, 3, 2, 1).

4 0

3 years ago

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Calculus

Differentiation

Basic Power Rule:

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$