So to start off you subtract y2 - y1/x2 - x1. This should look like -5 - -1/ -4 - -4. Now you do the math and you should get -4/0. Therefore the slope is undefined.
Answer:
he ded
Step-by-step explanation:
\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \tohe no alive because ⇆ω⇆π⊂∴∨α∈\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to![\neq \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n 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37,800 since it depreciates by 8% of the original price each year
Answer:
12x⁴ - 15x³ + 18x²
Step-by-step explanation:
1. distribute -3x²
-3x² • 5x = -15x³
-3x² • -4x² = 12x⁴
-3x² • -6 = 18x²
2. combine terms and put in descending order (standard form)
12x⁴ - 15x³ + 18x²
Answer: I believe the answer is (2) 49
Explanation: The given ratio is 5:245
When simplified(245/5), it gives you 1:49!