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olya-2409 [2.1K]
3 years ago
12

Explain why there are only three variables in the expression f(x) = a(b)^x + c as it relates to transformations of an exponentia

l function
Mathematics
1 answer:
UkoKoshka [18]3 years ago
7 0

Answer:

Option A.Exponential decay, 55% decrease

we have

This is a exponential function of the form

where a is the initial value

b is the base

b=(1+r)

r is the rate of change

In this problem

a=54

b=0.45

so

0.45=1+r

r=0.45-1

r=-0.55

Convert to percentage

r=-55% ------> is negative because is a exponential decay

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A teacher is giving each student 5/6 cup of trail mix. If there are 15 students in the class how much trail mix does the teacher
SCORPION-xisa [38]

15 * 5/6 cups = 75/6 cups

12  3/6 cups

12 1/2 cups of trail mix needed

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3 years ago
A complex number, represented by z = x + iy, may also be visualized as a 2 by 2 matrix
Marat540 [252]

Answer:

Step-by-step explanation:

A) Suppose that we have the complex numbers

z= x + iy \quad \text{and} \quad \\\\ \tilde{z}=\tilde{x} + i \tilde{y}

Remember that to sum complex numbers, we sum the real parts of the two numbers to get the real part and the imaginary parts of the two numbers to get the imaginary part. Hence,  

z+\tilde{z} = (x + i y) + (\tilde{x} + i \tilde{y}) = (x + \tilde{x})+i (y+\tilde{y})

On the other hand, if we sum the matrix visualizations of z \quad \text{and} \quad \tilde{z} we get

\left[\begin{array}{cc}x &y\\-y&x\end{array}\right] + \left[\begin{array}{cc}\tilde{x}&\tilde{y}\\ -\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x + \tilde{x}& y + \tilde{y}\\-(y+\tilde{y})&x+\tilde{x}\end{array}\right]

which is the matrix visualization of z + \tilde{z}.

To multiply two complex numbers, we use the distributive law to multiplly and then separete the real part from the imaginary part

z \cdot \tilde{z}= (x + iy) \cdot (\tilde{x} + i \tilde{y})=(x \tilde{x} + i x \tilde{y} + i \tilde{x} y - y\tilde{y} ) = (x\tilde{x}-y\yilde{y})+i(x\tilde{y}+\tilde{x}y)

Again, if we multiply the matrix visualizations of z \quad \text{and} \quad \tilde{z} we get

\left[\begin{array}{cc}x&y\\-y&x\end{array}\right]\left[\begin{array}{cc}\tilde{x}&\tilde{y}\\-\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x\tilde{x}-y\tilde{y}&x\tilde{y}+y\tilde{x}\\-y\tilde{x}-x\tilde{y}&x\tilde{x}-y\tilde{y}\end{array}\right]

which is the matrix viasualization of z\cdot\tilde{z}.

B)  Since the usual matrix operations are consisten with the usual addition and multiplication rules in the complex numbers, we can use them to find the multiplicative inverses of a complex number z=x+iy.

We are looking for the complex number z^{-1}=(x+iy)^{-1} which in terms of matrices is equivalent to find the matrix

\left[\begin{array}{cc}x&y\\-y & x\end{array}\right]^{-1}= \dfrac{1}{x^{2}+y^{2}} \left[\begin{array}{ccc}x&-y\\y&x\end{array}\right]    

Hence,

z^{-1}=\dfrac{1}{x^2 +y^2} (x-iy)=\dfrac{1}{|z|^2}(x-iy)

6 0
3 years ago
Simplify 8(5y-4x)/15 - ( - 3x + 7y/2 +3x - 4y)​
vekshin1

Answer:

that's it I think its clear

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If a tennis ball has a diameter of 14, What is the volume of the tennis ball in terms of pie?
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Answer:

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Step-by-step explanation:

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