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
(1, 1)
(3, 5)
(-1, 5)
Answer:
The point that is a solution of x + 2y ≤ 4, is (1,1)
Step-by-step explanation:
Given x + 2y ≤ 4, to teste if a point is a solution of the inequation you have to replace each par of values in the inequation and see the result.
So,
(2,4) ⇒ 2 + 2·4 ≤ 4 ⇒ 8 ≤ 4, this is not correct.
(1,1) ⇒ 1 + 2·1 ≤ 4 ⇒ 3 ≤ 4, is correct.
(3,5) ⇒ 3 + 2·5 ≤ 4 ⇒ 13 ≤ 4, this is not correct.
(-1,5) ⇒ -1 + 2·5 ≤ 4 ⇒ 9 ≤ 4, this is not correct.
Summarizing the only point that is solution of x + 2y ≤ 4, is (1,1)
3x - 8 = 15
Answer:x = 7.667
Step-by-step explanation:
3x - 8 = 15
+ 8 = 23
3x = 23
divide 3 by 3 and 23 so you get x by itself
answer = 7.667
Answer:
x = 7.667
Step-by-step explanation:
3x - 8 = 15 (Rearrange expression)
-8 - 15 = -3x (Combine like terms)
-23 = -3x (Divide)
x = 7.667
Answer:
1
Step-by-step explanation:
The question seems incomplete so rounding to the nearest whole number I’m guessing your answer is 1
The number of pay-per-view movies that would generate an equal monthly bill from both E-Z View cable company and View the Best company is 15 movies. This is found by setting up and solving a linear equation based on the monthly charges of each company.
This question is about using linear equations to solve a real-world problem. Let's say 'x' is the number of pay-per-view movies. Therefore, we can form two equations:
To find the number of pay-per-view movies that would make the two monthly costs equal, set the two equations equal to each other and solve for x.
#SPJ11
Answer:
Step-by-step explanation:
Answer-
The measurement of angle A' will be 130° .
Solution-
Transformations like - rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figure is not changed, then the figures are congruent.
It doesn't matter the order or how much degree of rotation has taken place, the final image will be congruent to the original image.
So, even after the rotation of 90° clockwise around the origin, the measurement of angle A will be 130° and so do all the rest of the angles.
Answer:
B, 130 degrees.
Step-by-step explanation: