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3 years ago

PLEASE HELP ASAP!! I WILL MARK BRAINIEST!! LOOK AT PHOTO

Mathematics

1 answer:

dusya [7]3 years ago

3 0

Answer:

True

Step-by-step explanation:

$\frac{5^{4} }{5^{2} } = \frac{5*5*5*5}{5*5} = 5*5 = 5^{2}$

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Which expression is equivalent to log Subscript c Baseline StartFraction x squared minus 1 Over 5 x EndFraction?

Harrizon [31]

The equivalent expression of $\log_c(\frac{x^2 - 1}{5x})$ is $\log_c(x^2 - 1) - \log_c(5x)$

<h3>How to determine the equivalent expression?</h3>

The logarithmic expression is given as:

$\log_c(\frac{x^2 - 1}{5x})$

The law of logarithm states that:

log(a) - log(b) = log(a/b)

This means that the expression can be split as:

$\log_c(\frac{x^2 - 1}{5x}) = \log_c(x^2 - 1) - \log_c(5x)$

Hence, the equivalent expression of $\log_c(\frac{x^2 - 1}{5x})$ is $\log_c(x^2 - 1) - \log_c(5x)$

Read more about equivalent expression

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2 years ago

Which of the functions have a range of all real numbers greater than or equal to 1 or less than or equal to -1? check all that a

kirill115 [55]

I believe it is A and D

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3 years ago

Read 2 more answers

A go carts too speed is 607200 feet per hour. what is the speed in mph?

densk [106]

The answer is 115 miles per hour

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4 years ago

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What is the area of the semicircle?

Serggg [28]

Step-by-step explanation:

$Area (semi-circle)= \frac{r^2\pi }{2}$

$A = \frac{(11)^2\pi }{2}$

$A=\frac{121\pi }{2}$

$A=\frac{379.94}{2}$

$A=189.97$

The area of the semi-circle is 189.97

6 0

3 years ago

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.5 parts

Hunter-Best [27]

Answer:

The value of the test statistic is $t = 2.1$

Step-by-step explanation:

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the expected value, $\sigma$ is the standard deviation and n is the size of the sample.

The level of ozone normally found is 7.5 parts/million (ppm).

This means that $\mu = 7.5$

The mean of 24 samples is 7.8 ppm with a standard deviation of 0.7.

This means that $X = 7.8, \sigma = 0.7, n = 24$

Test Statistic:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$t = \frac{7.8 - 7.5}{\frac{0.7}{\sqrt{24}}}$

$t = 2.1$

The value of the test statistic is $t = 2.1$

7 0

3 years ago