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podryga [215]
3 years ago
9

Simplify -

Mathematics
1 answer:
saveliy_v [14]3 years ago
6 0
C. <span>1/2^-6 x 2^8  (2^8 = 2*2*2*2*2*2*2*2 = 256)
    </span><span>1/2^-6 x 256  (when ever you do something to the power of a negative you do 1/something to the power of the number.An example is 2^-2 = 1/2^2)
</span><span>so 1/2^-6 = 1/1/2^6 or 32 (message me if you want me to explain how I got that)
    32 x 256
   <u><em> 8,192 <---- answer
</em></u><em /><em />

d. </span><span>-(-2)^0 x (-2)^3 x (-2)^-4 divide (-2)^3  (anything ^0 = 1)
</span>      <span>-(1)   x    -8   x (-2)^-4 divide   -8      (-2^-4 = 1/-2^4)</span><span>
      -1     x    -8   x  1/16  divide   -8       (simplify)
             8/16 </span>÷ -8
<span>            16/8 x -8
            <u>  -16 <-- answer</u>


e.  I do not know the answers. Sorry
</span><span>

</span>
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7+z / 2

z = 10

 replace the value of z

7 +10 /2

 use order of operations ( division comes before addition)

7+5 = 12

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1. Four hundred thousand thirty two and ten point one


2. Standard form is adding it out ( 945.35 ) = 900 + 40 + 5 + .30 + .05


3. Expanded form ( 73,890 )

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Answer:

  a) f(x)

  b) g(x)

  c) for all values of x > 1, g(x) has a greater rate of change

Step-by-step explanation:

a) The rate of change of f(x) is the x-coefficient: 4. The average rate of change of g(x) on an interval can be found by dividing the change by the interval width. Here, the width of the interval of interest is 1-0 = 1, so the average rate of change is g(1) -g(0) = 4-2 = 2.

On the interval [0, 1], f(x) has the greater rate of change.

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b) The interval [2, 3] also has a width of 1, so the rate of change of g(x) on that interval is g(3) -g(2) = 20 -10 = 10. This value is greater than 4, so ...

on the interval [2, 3], g(x) has the greater rate of change.

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c) Part (b) demonstrates that g(x) will have a greater rate of change on some intervals. In fact, for any interval whose center is greater than x=1, g(x) will have a greater rate of change than f(x).

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In calculus terms, the rate of change of g(x) is g'(x) = 4x. This will be greater than 4 for any x > 1. The graph shows the rate of change is 4 at x=1, and is higher for x > 1. The average rate of change will be greater than 4 on any interval whose center is greater than x=1.

We can figure the average rate of change on the interval [a, b] as ...

  m = (g(b) -g(a))/(b -a)

  m = ((2b² +2) -(2a² +2))/(b -a) = 2(b² -a²)/(b-a) = 2(b+a)

For the average rate of change to exceed 4, the sum of the ends of the interval must exceed 2, which is to say the midpoint must exceed 1.

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