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ladessa [460]
3 years ago
14

Simplify: (4xy)^0 please help

Mathematics
2 answers:
Mnenie [13.5K]3 years ago
7 0
Anything to the 0 power is equal to 1.  Since the whole equation is raised to the 0 power, the answer is 1.
katrin2010 [14]3 years ago
5 0
I believe that would be 1
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A new shopping mall records 150 total shoppers on their first day of business. Each day after that, the number of shoppers is 15
Wewaii [24]

The number of shoppers is an illustration of an exponential function.

The expression for the number of shoppers is: \mathbf{f(n) = 150(1.15)^n}

The given parameters are:

<em />\mathbf{a = 150}<em> -- the number of shoppers on the first day of business</em>

<em />\mathbf{r = 15\%}<em> --- the rate</em>

<em />

Because, the number of shoppers <em>increases each day</em>, the number of shopper on a certain day is:

\mathbf{f(n) = a \times (1 + r)^n}

Substitute 15% for r

\mathbf{f(n) = a \times (1 + 15\%)^n}

Express percentage as decimal

\mathbf{f(n) = a \times (1 + 0.15)^n}

\mathbf{f(n) = a \times (1.15)^n}

Substitute 150 for a

\mathbf{f(n) = 150 \times (1.15)^n}

\mathbf{f(n) = 150(1.15)^n}

Hence, the expression for the number of shoppers is: \mathbf{f(n) = 150(1.15)^n}

Read more about exponential functions at:

brainly.com/question/11487261

4 0
3 years ago
Need boys to add to a study group chat 9-12 grade plzzzz I already Have other students in it need more that go to connections/co
Elenna [48]

Answer:

So do the people who want to study with you give their number? If so, then I can't come cause I am a girl.

5 0
3 years ago
PLEASE HELP! In a film study, a sample of 300 movies was found to have an average runtime of 116 minutes with a standard deviati
BigorU [14]

Answer:

The answer is 1.42!!!!!!

Step-by-step explanation:

8 0
3 years ago
Remi invests £600 for 5 years in a savings account.
Daniel [21]

Answer:

225600

Step-by-step explanation:

Simple interest formula:

A = P(1 + rt)

A = final amount

P = initial principal balance

r = annual interest rate

t = time (in years)

So...

A = 600(1 +(75 x 5) )

75 x 5 = 375

1 + 375 = 376

600 x 376 = 225600

A = 225600

6 0
3 years ago
P(x) =x and q(x) = x-1Given:minimum x and Maximum x: -9.4 and 9.4minimum y and maximum y: -6.2 and 6.2Using the rational functio
Arte-miy333 [17]

we have the following function

\frac{p(x)}{g(x)}=\frac{x}{x\text{ -1}}

where x is between -9.4 and 9.4 and y is between -6.2 and 6.2.

We will first draw the function

from the graph, we can see that the zeroes are all values of x for which the graph crosses the x -axis

In this case, we see that that the only zero is at x=0.

Now, we have that the asymptotes are lines to which the graph of the function get really close to. On one side, we can see that as x goes to infinity or minus infinity, the values of the function get really close to 1. So the graph has a horizontal asymptote at y=1. Also, we can see that as x gets really close to 1, the graph gets really close to the vertical line x=1. So the graph has a vertical asymptote at x=1.

Recall that the domain of a function is the set of values of x for which the function is defined. From our graph, we can see that graph is not defined when x=1. So the domain of the function is the set of real numbers except x=1. Now, recall that the range of the function is the set of y values of the graph. From the picture we can see that the graph has a y coordinate for every value of y except for y=1. So, this means that the range of the function is the set of real numbers except y=1.

From the graph, we can see that we cannot draw the graph having a continous drawing. That is, imagine we take a pencil and start on one point on the graph on the left side. We can draw the whole graph on the left side, but we cannot draw the graph on the right side without lifting the pencil up. As we have to "lift the pencil up" this means that the graph is not continous

Finally note that as we have a vertical asymptote at x=1 and horizontal asymptote at y=1 we have that when y is 1 or x is 1, the function y=f(x)/g(x) is undefined

5 0
1 year ago
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