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Gre4nikov [31]
3 years ago
9

The curve the order pairs (0,10) and (1,5) and (2,2.5) can be represented by the function f(x)=10(0.5)x What is the multiplicati

ve rate of change if the functions
Mathematics
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

The multiplicative rate of change is 0.5.

Step-by-step explanation:

The given function is

f(x)=10(0.5)^x               .... (1)

It is given that the curve is passing through (0,10), (1,5) and (2,2.5).

The exponential function is defined as

f(x)=ab^x                .... (2)

Where, a is initial value and b is growth factor of multiplicative rate of change.

On comparing (1) and (2), we get

a=10

b=0.5

Therefore the multiplicative rate of change is 0.5.

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Rudiy27
Looking at this problem in terms of geometry makes it easier than trying to think of it algebraically.

If you want the largest possible x+y, it's equivalent to finding a rectangle with width x and length y that has the largest perimeter.

If you want the smallest possible x+y, it's equivalent to finding the rectangle with the smallest perimeter.

However, the area x*y must be constant and = 100.

We know that a square has the smallest perimeter to area ratio. This means that the smallest perimeter rectangle with area 100 is a square with side length 10. For this square, x+y = 20.

We also know that the further the rectangle stretches, the larger its perimeter to area ratio becomes. This means that a rectangle with side lengths 100 and 1 with an area of 100 has the largest perimeter. For this rectangle, x+y = 101.

So, the difference between the max and min values of x+y = 101 - 20 = 81.
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is the correct answer

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3 years ago
P is inversely proportional to the cube of (q-2) p=6 when q=3 find the value of p when q is 5
sveticcg [70]
\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
\\
&&y=\cfrac{{{  k}}}{x}
\end{array}\\\\
-----------------------------\\\\
\textit{p is inversely proportional to the cube of (q-2)}\implies p=\cfrac{k}{(q-2)^3}
\\\\\\
now \quad 
\begin{cases}
p=6\\
q=3
\end{cases}\implies 6=\cfrac{k}{(3-2)^3}

solve for "k", to find k or the "constant of variation"

then plug k's value back to \bf p=\cfrac{k}{(q-2)^3}

now.... what is "p" when q = 5?  well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is
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