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garri49 [273]
3 years ago
11

If you hade to use Algebar tiles to do the charge method subtraction and you problème is -9 - (-2)

Mathematics
2 answers:
Bas_tet [7]3 years ago
7 0

Answer:

-7

Step-by-step explanation:

Well the subtraction problem is now going to be a adding and the (-2) is now positive 2 so I'm thinking the answere is -7

gizmo_the_mogwai [7]3 years ago
6 0

Answer:

-9 - (-2)= -7

Step-by-step explanation:

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kompoz [17]

I would help but I cannot read it.

3 0
3 years ago
9+x/15=10<br><br> What is happening to the variable? <br><br><br> How do you solve for the variable?
Katarina [22]

Answer:

The variable x is being divided by 15 in the equation and x=15

Step-by-step explanation:

To solve for x we :

9+x/15=10

x/15=1

x=15

4 0
3 years ago
Find the derivative.
krek1111 [17]

Answer:

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

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients

  • Expanding/Factoring

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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:

<u>Step 1: Define</u>

<em>Identify</em>

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

<u>Step 2: Differentiate</u>

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

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

Unit: Differentiation

7 0
2 years ago
The x co-ordinate of any point lying on the Y axis is: please help urgent​
olchik [2.2K]

Answer:

x-coordinate = 0

Step-by-step explanation:

In a coordinate plan there are two perpendicular lines one is x-axis and another is y-axis

x-axis is a horizontal line and y-axis is a vertical line.

Both lines intersect each other at (0,0).

On x-axis, y-coordinate remains same , i.e., y=0.

On y-axis, x-coordinate remains same , i.e., x=0.

Therefore, the x co-ordinate of any point lying on the y axis is 0.

8 0
3 years ago
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