What is the area of a circle with a diameter of 12 m?

Which products result in a difference of squares? Check all that apply.

Amanda [17]

Answer:

2nd - (w - 5)(w + 5)

4th - (-4v - 9)(-4v + 9)

Step-by-step explanation:

1. The first option shows an expression multiplied by its opposite(x -1), so therefore, it does not show the difference of squares

2. The second option does show the difference of squares because it is in the form (a + b)(a - b)

3. The third option is just a square because the same expression is multiplied by itself.

4. The fourth option is the difference of squares because it is in the form (a + b)(a - b). a equals -4v and b equals 9 in this case.

5. The fifth option is not the difference of squares. No term in common in both expressions

6. The sixth option is just a square because the same expression is multiplied by itself.

In all, there are two options that are the difference of squares, the 2nd and 4th.

5 0

3 years ago

Ann lives 3/10 mile from school. Carlos lives 1/10 mile from school. How much farther from school does Ann live than Carlos?

sattari [20]

3/10 - 1/10 =2/10
Thats what I think

7 0

2 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago

At Edwards Middle School there are 8 teachers for every 136 students. There are 850 students enrolled at the school. How many te

In-s [12.5K]

Answer:

50

Step-by-step explanation:

one teacher per 17 students

7 0

2 years ago

Read 2 more answers

Find x and y part 4 for these problems

Dvinal [7]

Answer:

Step-by-step explanation:

3) Sin30 = 11/x

x = 11/Sin30 = 11/0.5

x = 22

Tan 30 = 11/y

y = 11/tan30 = 11/0.5774

y = 19.1

4) Sin30 = 6/x

x = 6/Sin30 = 6/0.5

x = 12

Tan 30 = 6/y

y = 6/tan30 = 6/0.5774

y = 10.39

5) Sin45 = 9√2/y

y = 9√2/Sin45 = 9√2/(√2/2) =

9√2 × 2/√2 = 18

x = 18

Tan 45 = 9√2/x

x = 9√2/Tan 45 = 9√2/1

x = 9√2

6)

Sin60 = 9/x

x = 9/Sin60 = 9/0.866

x = 10.39

Tan 60 = 9/y/2 = 18/y

1.7321 = 18/y

y = 18/1.7321

y = 10.39

6 0

2 years ago