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3 years ago

Write X^-2y^0z^2/xy using only positive exponents

Mathematics

1 answer:

iogann1982 [59]3 years ago

3 0

Answer:

z^2/(x^3y)

Step-by-step explanation:

We assume you want to simplify ...

$\dfrac{x^{-2}y^0z^2}{xy}=x^{-2-1}y^{0-1}z^2=x^{-3}y^{-1}z^2\\\\=\boxed{\dfrac{z^2}{x^3y}}$

_____

Comment on the given expression

What you have written is ...

(x^-2y^0z^2/x)y = yz^2/x^3

The order of operations requires that division and multiplication be done in the order of appearance, so a/xy means (a/x)y unless you put parentheses on the denominator product. Typesetting the expression lets you use the fraction bar for grouping, so parentheses are not needed when that form is used.

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I'd seriously appreciate it if anyone could help ASAP! :) I'll give Brainliest!

Jet001 [13]

We have been given a graph of function g(x) which is a transformation of the function $f(x)=3^x$

Now we have to find the equation of g(x)

Usually transformation involves shifting or stretching so we can use the graph to identify the transformation.

First you should check the graph of $f(x)=3^x$

You will notice that it is always above x-axis (equation is x=0). Because x-axis acts as horizontal asymptote.

Now the given graph has asymptote at x=-2

which is just 2 unit down from the original asymptote x=0

so that means we need shift f(x), 2 unit down hence we get:

$y=3^x$

but that will disturb the y-intercept (0,1)

if we multiply $3^x$ by 3 again then the y-intercept will remain (0,1)

Hence final equation for g(x) will be:

$g(x)=3(3^x)-2$

6 0

3 years ago

Xavier opened a savings account with a deposit of $12,000. The account earned simple interest. He did not make any additional de

klasskru [66]

The annual interest rate on the saving account of Xavier which is earned simple interest is 9%.

<h3>What is simple interest?</h3>

Simple interest is the amount charged on the principal amount with a fixed rate of interest for a time period. Simple interest calculated only on the principal amount.

The formula for the simple interest can be given as,

$I=\dfrac{PRT}{100}$

Here, (I) is the interest amount on the principal amount of (P) with the rate of (r) in the time period of (t).

Xavier opened a savings account with a deposit of $12,000. At the end of 3 years, the account balance was $12,324.

The principal amount is $12000 and the total amount is $12324. Thus, the interest earned by the Xavier is,

$I=12324-12000\\I=324$

Total time period is 3 years and the account earned simple interest. Thus, put these values in the above formula as,

$324=\dfrac{(12000)R(3)}{100}\\R=\dfrac{32400}{12000\times3}\\R=9\%$

Hence, the annual interest rate on the saving account of Xavier which is earned simple interest is 9%.

Learn more about the simple interest here;

brainly.com/question/2294792

8 0

2 years ago

Which pair of numbers has the greatest common factor As that of 48 and 78 ?

Nataly_w [17]

48 and 78(try to see if its 6)

5 0

3 years ago

Tell whether the statement is true or false. If w is 25% of Z, then Z:W is 75:25

melamori03 [73]

False.
If w is 25% of z, then w:z would be 25:100, because 25 is 25% of 100.
If Z:W was 75:25, it would be wrong because 25 is 33% of 75, not 25%.

7 0

3 years ago

The weight of National Football League (NFL) players has increased steadily, gaining up to 1.5 lb. per year since 1942. Accordin

GuDViN [60]

Answer:

The probability that the sample mean weight will be more than 262 lb is 0.0047.

Step-by-step explanation:

The random variable X can be defined as the weight of National Football League (NFL) players now.

The mean weight is, μ = 252.8 lb.

The standard deviation of the weights is, σ = 25 lb.

A random sample of n = 50 NFL players are selected.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The sample of players selected is quite large, i.e. n = 50 > 30, so the central limit theorem can be used to approximate the distribution of sample means.

$\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)$

Compute the probability that the sample mean weight will be more than 262 lb as follows:

$P(\bar X>262)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{262-252.8}{3.536})\\\\=P(Z>2.60)\\\\=1-P(Z$

*Use a z-table for the probability.

Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.

6 0

3 years ago