This question is already answered......
Answer:
x = 6
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
- Subtract Property of Equality
<u>Algebra I</u>
- Terms/Coefficients/Degrees
- Expand by FOIL (First Outside Inside Last)
- Factoring
- Multiple Roots
<u>Trigonometry</u>
[Right Triangles Only] Pythagorean Theorem: a² + b² = c²
- a is a leg
- b is another leg
- c is the hypotenuse
Step-by-step explanation:
<u>Step 1: Identify</u>
<em>a</em> = x + 3
<em>b</em> = x
<em>c</em> = √117
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute [PT]: (x + 3)² + x² = (√117)²
- Expand [FOIL]: x² + 6x + 9 + x² = (√117)²
- Combine like terms: 2x² + 6x + 9 = (√117)²
- Exponents: 2x² + 6x + 9 = 117
- [SPE] Subtract 117 on both sides: 2x² + 6x - 108 = 0
- Factor out GCF: 2(x² + 3x - 54) = 0
- [DPE] Divide 2 on both sides: x² + 3x - 54 = 0
- Factor Quadratic: (x - 6)(x + 9) = 0
- Solve roots/solve <em>x</em>: x = -9, 6
Since we are dealing with positive values, we can disregard the negative root.
∴ x = 6
You can calculate it using the law of cosines: c^2=a^2+b^2-2*a*b*cos(C)
your triangle is
CD=15=a
CE=?=b
DE=CE+3=b+3=c
and C=90°
-> insert those values, with c substituted with b+3 to remove c
c^2=a^2+b^2-2*a*b*cos(C)
(b+3)^2=15^2+b^2-2*15*b*cos(90)
cos(90)=0->
(b+3)^2=15^2+b^2
b^2+2*3*b+3^2=225+b^2
6b+9=225
6b=216
b=36=CE
DE=CE+3=36+3=39
