Answer:
The solution is any point in the common part of red and blue
The two lines intersect each other at (4 , 7)
Step-by-step explanation:
∵ y > 1/4 x + 6 ⇒ y = 1/4 x + 6
∵ y > 2x - 1 ⇒ y = 2x - 1
∴ 2x - 1 = 1/4 x + 6
∴ 2x - 1/4 x = 6 + 1
∴ 7/4 x = 7 ⇒ x = 7 ÷ 7/4
∴ x = 4
∴ y = 2(4) - 1 = 7
∴ The two lines intersect each other at (4 , 7)
∴ The solution is any point in the common part of red and blue
They would be similar triangles because they cannot be congruent due to the fact that the side lengths are different. They are dilutions of each other which again proves that they are similar, not congruent.
Answer:
The 1st option
Step-by-step explanation:
Notice in sequence 1, it is 1,6; 2,12
That is sequence 1 * 6 = sequence 2.
Hope this helps! Have a nice day!
Answer:
Intercepts:
x = 0, y = 0
x = 1.77, y = 0
x = 2.51, y = 0
Critical points:
x = 1.25, y = 4
x = 2.17
, y = -4
x = 2.8, y = 4
Inflection points:
x = 0.81, y = 2.44
x = 1.81, y = -0.54
x = 2.52, y = 0.27
Step-by-step explanation:
We can find the intercept by setting f(x) = 0
where n = 0, 1, 2,3, 4, 5,...
Since we are restricting x between 0 and 3 we can stop at n = 2
So the function f(x) intercepts at y = 0 and x:
x = 0
x = 1.77
x = 2.51
The critical points occur at the first derivative = 0
or
where n = 0, 1, 2, 3
Since we are restricting x between 0 and 3 we can stop at n = 2
So our critical points are at
x = 1.25,
x = 2.17
,
x = 2.8,
For the inflection point, we can take the 2nd derivative and set it to 0
We can solve this numerically to get the inflection points are at
x = 0.81,
x = 1.81,
x = 2.52,
468
You can get this by finding the area of each individual rectangle.