Answer:
It is A: 11 to 1 :)
It's true!
(edited my answer btw, you have to simply also welp)
Answer:
Step-by-step explanation:
Actually Welcome to the Concept of the linear equations..
Here given value of x= 5 and y =1 , so we get as,
5a + b = 38 and 5b - a = 8
so, now we multiply equation no. 2 by 5 all over.
==> 25b - 5a = 40....(1)
hence adding new equation and equation no. 1
26b = 78
b = 78/26
hence b = 3 , and a = 7
Answer:
C
Step-by-step explanation:
The minimum you can stay is 1 night, which would be 100 dollars. Since there's a fee, and can only stay 7 nights, the max number would be 900.
The given function are
r(x) = 2 - x² and w(x) = x - 2
<span>(w*r)(x) can be obtained by multiplying the both function together
</span>
So, <span>(w*r)(x) = w(x) * r(x) = (x-2)*(2-x²)</span>
<span>(w*r)(x) = x (2-x²) - 2(2-x²)</span>
= 2x - x³ - 4 + 2x²
∴ <span>
(w*r)(x) = -x³ + 2x² + 2x - 4</span>
<span>It is a polynomial function with a domain equal to R
</span>
The range of <span>(w*r)(x) can be obtained by graphing the function
</span>
To graph (w*r)(x), we need to make a table between x and (w*r)(x)
See the attached figure which represents the table and the graph of <span>(w*r)(x)
</span>
As shown in the graph the range of <span>
(w*r)(x) is (-∞,∞)</span>
<span>The graph you plotted is the graph of f ' (x) and NOT f(x) itself. </span>
Draw a number line. On the number line plot x = 3 and x = 4. These values make f ' (x) equal to zero. Pick a value to the left of x = 3, say x = 0. Plug in x = 0 into the derivative function to get
f ' (x) = (x-4)(6-2x)
f ' (0) = (0-4)(6-2*0)
f ' (0) = -24
So the function is decreasing on the interval to the left of x = 3. Now plug in a value between 3 and 4, say x = 3.5
<span>f ' (x) = (x-4)(6-2x)
</span><span>f ' (3.5) = (3.5-4)(6-2*3.5)
</span>f ' (3.5) = 0.5
The function is increasing on the interval 3 < x < 4. The junction where it changes from decreasing to increasing is at x = 3. This is where the min happens.
So the final answer is C) 3