The equivalence of the equation is 5x + 48 = 5x + 3.
Since 48 cannot be equal to 3, hence there are no solution.
<h3>What is the value of x?</h3>
Given the equation; x-4[ x - 2( x + 6) ] = 5x + 3
We remove the parentheses
x-4[ x - 2( x + 6) ] = 5x + 3
x-4[ x - 2x - 12 ] = 5x + 3
x - 4x + 8x + 48 = 5x + 3
5x + 48 = 5x + 3
We can go further and collect like terms
5x - 5x + 48 = 3
48 ≠ 3
Since 48 cannot be equal to 3, hence there are no solution.
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AB²=5²+5²+5²=25+25+25=75
AB=5√3≈8.7
Answer:AB≈8.7
Answer:
(2, 4)
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>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = 2x
x = -y + 6
<u>Step 2: Solve for </u><em><u>y</u></em>
<em>Substitution</em>
- Substitute in <em>x</em>: y = 2(-y + 6)
- Distribute 2: y = -2y + 12
- Isolate <em>y</em> terms: 3y = 12
- Isolate <em>y</em>: y = 4
<u>Step 3: Solve for </u><em><u>x</u></em>
- Define equation: x = -y + 6
- Substitute in <em>y</em>: x = -4 + 6
- Add: x = 2
Answer: 129
Step-by-step explanation: First find the area of the circle. Since we know the area of a circle is πr² (pi multiplied by the radius squared) we can find the area of the half circle. The radius is 5 (because radius is half of the diameter)
Area=3.14(5)²
=3.14(25)
=78.5
But since this is just a half circle, divide this by 2
That will get 39.25
Then add this to the area of the rectangle (L*W) which is 90
That will get about 129
Answer:
Step-by-step explanation:
Insert the given x values for the given equation to find the corresponding y coordinate:
Insert -6:
Simplify multiplication (when multiplying a negative and a positive, the result will always be negative):
Simplify (when subtracting from a negative number, add the values, but keep the negative symbol):
Insert -1:
Insert 1:
:Done
