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skelet666 [1.2K]
2 years ago
10

Need asaaaap

Advanced Placement (AP)
1 answer:
dem82 [27]2 years ago
7 0

Answer:

A mousetrap.

Explanation:

An irreducibly complex system is one of which all parts are necessary for it to function, so when only one part does not work, the whole system does not work. Thus, it is necessary that all parts work in a complementary way, in order to achieve a positive result. An example is the mouse trap: remove an item from all parts and the trap stops working.

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I need to find Y, process?
shepuryov [24]

Answer:

If you look at Y and compare it to 120, Y looks as if it is half of what the angle measurement 120 is. So you would do 120 divided by 2 (to get half of 120) and you end up with 60. We can infer that the top left angle measures 60. So, you do 60 divided by 5 (since you are looking for Y but there is a 5 in front of it so you would multiply 5 and Y to get 60. You need to do the inverse operation which is divide.) So you divide 60 and 5 and get 12. Replace Y with 12 and you have 5(12). The answer is Y=12

Explanation:

Hope this helps!

7 0
2 years ago
My fellow math brodas, help
DiKsa [7]

A. Depending on which variable you choose to integrate with, you can capture the total bounded region with either -2 ≤ x ≤ (-1 + √5)/2 or 1 ≤ y ≤ (5 + √5)/2, where the upper endpoints correspond to the coordinates of the appropriate intersections:

y = x² + 1

⇒   x = (x² - 2)² - 2

⇒   x⁴ - 4x² - x + 2 = 0

⇒   (x - 2) (x + 1) (x² + x - 1) = 0

⇒   x = 2, x = -1, x = -1/2 ± √5/2

⇒   y = 5, y = 2, y = (5 ± √5)/2

On the other hand, we can compute the areas of A and B separately, then sum those integrals. Area A is easier to compute by integrating with respect to y over 2 ≤ y ≤ (5 + √5)/2, while area B is easier to find by integrating x over -1 ≤ x ≤ (-1 + √5)/2.

B. I'll stick to the split-region approach.

First, we find equations for the appropriates halves of either parabola:

• y = x² + 1   ⇒   x = ± √(y - 1)

and x = -√(y - 1) describes the left half of the blue parabola;

• x = (y - 3)² - 2   ⇒   y = 3 ± √(x + 2)

and y = 3 - √(x + 2) describe the bottom half of the red parabola.

Now we can set up the integrals.

Area of A:

\displaystyle \int_2^{(5+\sqrt5)/2} \left(\left(-\sqrt{y-1}\right) - \left((y-3)^2-2\right)\right) \, dy \\ ~~~~~~~~ = -\int_2^{(5+\sqrt5)/2} \left((y-3)^2 - 2 + \sqrt{y-1}\right) \, dy

Area of B:

\displaystyle \int_{-1}^{(-1+\sqrt5)/2} \left(\left(3-\sqrt{x+2}\right) - \left(x^2+1\right) \right) \, dx \\ ~~~~~~~~ = - \int_{-1}^{(-1+\sqrt5)/2} \left(x^2 - 2 + \sqrt{x+2}\right) \, dx

Alternatively, one can prove that the regions A and B are symmetric across the line y = x + 3, so we can simply pick one of these integrals and double it.

C. Computing the integrals, we find

area of A = 2/3

area of B = 2/3

and so the total area is 2/3 + 2/3 = 4/3.

6 0
1 year ago
I need help with these, there are already filled in just if anyone knows these off the top of their head it would be greatly app
slava [35]

Your descriptors are mostly correct, assuming the column headings are something like ...

... function ... domain ... range ... odd/even ... intercepts ... increasing/decreasing ... ??? ... critical points

_____

√x is increasing on the interval [0, ∞). There is no branch where the function is decreasing.

1/x has a critical point at x=0. The slope is undefined there.

3 0
3 years ago
What concentrations is chloroform sold in
eimsori [14]
.concentration range 40,000 ppm
6 0
3 years ago
President Franklin Roosevelt had issues with the Supreme Court when it found some of his programs unconstitutional. In an attemp
Lunna [17]

Answer:  the first election returns reached his family estate in Hyde Park, New York, on a November night in 1936, Franklin Delano Roosevelt leaned back in his wheelchair, his signature cigarette holder at a cocky angle, blew a smoke ring and cried “Wow!” His huge margin in New Haven signaled that he was being swept into a second term in the White House with the largest popular vote in history at the time and the best showing in the electoral college since James Monroe ran unopposed in 1820.

The outpouring of millions of ballots for the Democratic ticket reflected the enormous admiration for what FDR had achieved in less than four years. He had been inaugurated in March 1933 during perilous times—one-third of the workforce jobless, industry all but paralyzed, farmers desperate, most of the banks shut down—and in his first 100 days he had put through a series of measures that lifted the nation’s spirits. In 1933 workers and businessmen marched in spectacular parades to demonstrate their support for the National Recovery Administration (NRA), Roosevelt’s agency for industrial mobilization, symbolized by its emblem, the blue eagle. Farmers were grateful for government subsidies dispensed by the newly created Agricultural Adjustment Administration

8 0
3 years ago
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