(3x^3 + 4x²) + (3x^3 – 4x^2 – 9x) =
We have to combine like terms, terms that have x raised to the same power:
First, eliminate the parenthesis:
3x^3 + 4x² + 3x^3 – 4x^2 – 9x =
We have 2 terms raised to the second power (4x^2 and -4 x^2) simply add and subtract.
3x^3+3x^3+4x^2-4x^2-9x
6x^3-9x
We have that
x²<span>-6x+7=0
</span>Group terms that contain the same variable
(x²-6x)+7=0
Complete the square Remember to balance the equation
(x²-6x+9-9)+7=0
Rewrite as perfect squares
(x-3)²+7-9=0
(x-3)²-2=0
(x-3)²=2
(x-3)=(+/-)√2
x=(+/-)√2+3
the solutions are
x=√2+3
x=-√2+3
Answer:
its B the answer is b
Step-by-step explanation:
Answer:
- The center (2, 2.5), radius
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Step-by-step explanation:
<u>The standard form of the equation of a circle is: </u>
- ( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius
<u>Rewrite the given equation in the standard form:</u>
- 2x^2 + 2y^2 - 8x + 10y + 2 = 0
- x^2 - 4x + y^2 + 5y = -1
- x^2 - 4x + 2^2 + y^2 + 5y + (5/2)^2 = -1 + 4 + 25/4
- (x - 2)^2 + (y + 2.5)^2 = 37/4
<u>The center is:</u>
<u>And radius is:</u>
- <u />
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Answer:
(a) If r is any rational number and s is any irrational number, then r/s is rational
(b) The statement is false when r is 0
Step-by-step explanation:
Given
rational number
irrational number
irrational number
Solving (a): The negation
To get the negation of a statement, we only need to negate the end result
In other words, the number type of r and s will remain the same, but r/s will be negated.
So, the negation is:
rational number
irrational number
rational number
Solving (b): When r/s is irrational is false
Given that:
irrational number
Set r to 0
So:

-- rational
<em>Hence, the statement is false when r is 0</em>