Since you keep proportional dimensions, the proportion between the old and new dimensions must be the same. So, if we call the new height
, the preservation of the width/height ratio is written as

Solving the proportion for
yields

Answer:
B. x = -1 ± i
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Algebra I</u>
- Standard Form: ax² + bx + c = 0
- Factoring
- Quadratic Formula:

<u>Algebra II</u>
- Imaginary Numbers: √-1 = i
Step-by-step explanation:
<u>Step 1: Define</u>
x² + 2x = -2
<u>Step 2: Identify Variables</u>
- Rewrite Quadratic in Standard Form [Addition Property of Equality]: x² + 2x + 2 = 0
- Break up Quadratic: a = 1, b = 2, c = 2
<u>Step 3: Solve for </u><em><u>x</u></em>
- Substitute in variables [Quadratic Formula]:

- [√Radical] Evaluate exponents:

- Multiply:

- [√Radical] Subtract:

- [√Radical] Factor:

- [√Radicals] Simplify:

- Factor:

- Divide:

Answer:
(You never showed the graphs we were suppose to choose from)
The graphs of all exponential functions have these characteristics. They all contain the point (0, 1), because a0 = 1. The x-axis is always an asymptote. They are decreasing if 0 < a < 1, and increasing if 1 < a.
Answer:
A
Step-by-step explanation:
I added 6.5 to the equation, found the mean, and it was 3.5.
Understanding the Absolute Value.
First, know what the absolute value is.
The absolute value is the value that determines how far the value is from 0.
For example, The absolute value of -5 is far from 0 5 units. Therefore the absolute value of -5 equals 5.
Basic Absolute Value Defines
| a | = a
- | a | = -a
| - a | = a
Back to the question. To evaluate those expressions, we use the defines of absolute value.
|-16| = 16
|-1| = 1
16-(1)
Then remove the brackets. 16 - 1 = 15
Therefore, the answer is 15.
<em>The</em><em> </em><em>answer</em><em> </em><em>above</em><em> </em><em>is</em><em> </em><em>when</em><em> </em><em>being</em><em> </em><em>subtracted</em><em> </em><em>and</em><em> </em><em>evaluated</em><em> </em><em>from</em><em> </em><em>both</em><em> </em><em>16</em><em>-</em><em>1</em>
<em>Evaluating</em><em> </em><em>for</em><em> </em><em>each</em><em> </em><em>expressions</em><em> </em><em>would</em><em> </em><em>be</em>
<em>|</em><em>-</em><em>16</em><em>|</em><em> </em><em>=</em><em> </em><em>16</em>
<em>-</em><em>|</em><em>-</em><em>1</em><em>|</em><em> </em><em>=</em><em> </em><em>-</em><em>1</em>