Answer:
Step-by-step explanation:
x![\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left \{ {{y=2} \atop {x=2}} \right. \int\limits^a_b {x} \, dx](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx)
XE[_4.2,2.2]
Answer:
3(4x - 1)(2x + 3)
Step-by-step explanation:
Rearrange the equation into standard form
Subtract 9 - 30x from both sides
24x² + 30x - 9 = 0 ← in standard form
Take out 3 as a common factor
3(8x² + 10x - 3) = 0 ← factor the quadratic
Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x term
product = 8 × - 3 = - 24, sum = 10
The factors are - 2 and + 12
Use these factors to replace the x- term, that is
8x² - 2x + 12x - 3 ( factor the first/second and third/fourth terms )
2x(4x - 1) + 3(4x - 1) ← take out the common factor (4x - 1)
(4x - 1)(2x + 3)
24x² + 30x - 9 = 3(4x - 1)(2x + 3) ← in factored form
Let the number be x
Then the absolute value of the number from -2 should be 4
So either
Or
[tex] <B>The numbers are 2 and -6</B> [/tex[
They are four units away from -2 on the number line
The given distance of 5.20 would be A.
Replace A with 5.20 to solve for T.
T = 5.20^3/2
T = 11.9 years.
