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4 years ago

Which expressions are monomials?

Mathematics

2 answers:

noname [10]4 years ago

5 0

Answer:

A,C,E

Step-by-step explanation:

Nimfa-mama [501]4 years ago

4 0

Monomials are the expressions with one terms.

A.) $-4+6 = 2$

Since, it has only one term. Therefore, it is a monomial.

B.) $b+2b+2 = 3b+2$

Since, it has two terms. Therefore, it is not a monomial.

C.) $(x-2x)^2 = (-x)^2=x^2$

Since, it has only one term. Therefore, it is a monomial.

D.) $\frac{rs}{2}$

Since, it has only one term. Therefore, it is a monomial.

E.) $36x^2yz^3$

Since, it has only one term. Therefore, it is a monomial.

F.) $a^x$

Since, the variable is in the power. So it is not a monomial.

G.) $x^\frac{1}{3}$

Since, the power is not a integer. Therefore, it is not a monomial.

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4 years ago

Consider a circle whose size can vary. The circumference of the circle is always 2 π 2π times as large as its radius. Let r r re

Alenkasestr [34]

Answer: $C=2\pi r$

Step-by-step explanation:

Let be "C" the circumference of the circle (in feet) and "r" the radius of the circle (in feet).

Based on the information provided in the problem, you know that the circumference of the circle is always $2\pi$ as large as its radius.

Notice that this indicates a multiplication. Then, this means that the circumference of the circle is always equal to $2\pi$ by "r".

Based on this, you can write the following formula that expresses the circumference "C" in terms of the radius "r":

$C=2\pi r$

4 0

3 years ago

A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th

Burka [1]

Answer:

Confidence interval : $21.506$ to $24.493$

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = $\bar{x}=\frac{\sum x}{n}$

Sample mean = $\bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}$

Sample mean = $\bar{x}=23$

Sample standard deviation = $\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$

Sample standard deviation = $\sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}$