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Pepsi [2]
3 years ago
14

Which graph represents this system of equations? y + 2x = 3 y + 2 = 3x

Mathematics
1 answer:
andrezito [222]3 years ago
7 0

Answer:

The options are not given, so i will answer in a general way.

First, we have the system:

y + 2*x = 3

y + 2 = 3*x

To graph this system, the easier way is writing both equations as linear equations, like:

y = -2*x + 3

y = 3*x - 2

Now we just need to graph these two lines in the same coordinate axis.

To do it, we can replace the value of x by two different values in each equation, and see the correspondent value of y for these values. Then graph the two points, and draw a line that connects them.

For example, for the first equation if we have x = 0, we have:

y = -2*0 + 3

y = 3

Then we have the point (0, 3)

and if x = 1

y = -2*1 + 3 = 1

y = 1

we have the point (1, 1)

Now we only need to graph these points, and draw a line that passes through them.

Once we did this for both lines, the graph we will get will look something like the one above.

And the point where the lines intersect is the solution of the system of equations.

Which is the point (1, 1)

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The data in the table are linear. Use the table to find the slope.
Gemiola [76]

Answer:

m = -\frac{3}{2}

Step-by-step explanation:

Given

x 2 4 6 8

y 1 -2 -5 -8

Required

Determine the slope

First, pick any two corresponding x and y values.

(x_1,y_1) =(2,1)

(x_2,y_2) =(8,-8)\\

The slope (m) is then calculated using:

m = \frac{y_1 - y_2}{x_1 - x_2}

Substitute in, the x and y values

m = \frac{1 - (-8)}{2- 8}

m = \frac{1 +8}{-6}

m = \frac{9}{-6}

m = -\frac{9}{6}

Divide numerator and denominator by 3

m = -\frac{9/3}{6/3}

m = -\frac{3}{2}

<em>Hence, the slope is -3/2</em>

8 0
3 years ago
What is the derivative of 1/square root 4x.
Bumek [7]

Answer:

\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

Exponential Properties

  • Exponential Property [Rewrite]:                                                                   \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Property [Root Rewrite]:                                                           \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)  

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg]

<u>Step 2: Differentiate</u>

  1. Simplify:                                                                                                         \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \bigg( \frac{1}{2\sqrt{x}} \bigg)'
  2. Rewrite [Derivative Property - Multiplied Constant]:                                   \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{\sqrt{x}} \bigg)'
  3. Rewrite [Exponential Rule - Root Rewrite]:                                                 \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{x^\Big{\frac{1}{2}}} \bigg)'
  4. Rewrite [Exponential Rule - Rewrite]:                                                           \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( x^\bigg{\frac{-1}{2}} \bigg)'
  5. Derivative Rule [Basic Power Rule]:                                                             \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{-1}{2} x^\bigg{\frac{-3}{2}} \bigg)
  6. Simplify:                                                                                                         \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4} x^\bigg{\frac{-3}{2}}
  7. Rewrite [Exponential Rule - Rewrite]:                                                           \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}

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

Unit: Differentiation

5 0
3 years ago
I need help with part B
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