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alexandr1967 [171]
3 years ago
11

Eeek!! I don’t get this!

Mathematics
2 answers:
Neporo4naja [7]3 years ago
5 0

If the area is 1 sq meter, then that means that each of the smaller squares are 1/9 meters long and wide, since the square root of 1 is 1. To find the area of a right triangle, use the formula a*b/2. This makes 1/18 square meters for each triangle.

A.) 1/18 * 6 =  6/18 or 1/3 sq. meters

B.) The mini squares in this problem are made up of two right triangles, so the area of a mini square is twice that of a triangle. If we have nine mini squares, we have 18 triangles, which makes the 1 * 1 single square. What you should do is add half the area of a single square to the original:

18/2 = 9

So we add 9 to the eighteen right triangles, which means 27 triangles to make 1 1/2 square meters.

Hope this helped.

julsineya [31]3 years ago
3 0

Wow. They worked hard to trip you up on this one!

First, don't let the word "decompose" spook you.

We are basically dividing 1 sq m into 9 equal pieces.

So each piece is 1/9 of a sq m.

Then, we make each small piece 1/2 of a square.

1/9 x 1/2 = 1/18.

Each triangle will be 1/18 of a sq meter.

So 6 will be 6/18 or 1/3 of a sq. meter

To figure 1.5 square meters out of triangles...

There are 2 triangles per small square, and 9 small squares per sq meter - and then you add 50% more.

(2 x 9) + (2 x 9 x .5) = 27 triangles

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a = 2 and b = -4

============================================================

Explanation:

Let's define the three helper functions

  • f(x) = ax^2 - b
  • h(x) = 6
  • j(x) = 5ax+b

which are drawn from the piecewise function. The g(x) function will change depending on what the input is.

  • If x < 1, then g(x) = f(x).
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  • If x > 1, then g(x) = j(x)

Since we want g(x) to be continuous at x = 1, this must mean the three functions f(x), h(x), j(x) must have the same output value when the input is x = 1.

Because h(x) = 6 is a constant function, the output is always 6 regardless of the input. Therefore, we want f(x) and j(x) to have 6 as their output when x = 1. Or else, the pieces won't connect.

Plug x = 1 into the f(x) function to get

f(x) = ax^2 - b

f(1) = a(1)^2 - b

f(1) = a - b

Set this equal to the desired output of 6 and we end up with the equation a-b = 6. Solving for 'a' leads to a = b+6.

------------

We'll use the same idea for j(x)

j(x) = 5ax + b

j(1) = 5a(1) + b

j(1) = 5a + b

5a+b = 6

5(b+6) + b = 6 ... plug in a = b+6; solve for b

5b+30+b = 6

6b+30 = 6

6b = 6-30

6b = -24

b = -24/6

b = -4

Which then leads to,

a = b+6

a = -4+6

a = 2

------------

Since a = 2 and b = -4, we go from this

g(x) = \begin{cases}ax^2-b, \ \ x < 1\\6, \ \ x = 1\\5ax+b, \ \ x > 1\end{cases}

to this

g(x) = \begin{cases}2x^2+4, \ \ x < 1\\6, \ \ x = 1\\10x-4, \ \ x > 1\end{cases}

Meaning

f(x) = 2x^2+4 and j(x) = 10x-4

You should find that plugging x = 1 into each of those two functions leads to 6 as the output.

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