All categories

3 years ago

the end points of a diameter of a circle are A(3,1) and B(8,13). Find the area of the circle in terms of X

Mathematics

2 answers:

katrin2010 [14]3 years ago

7 0

Answer:

I think it's 4.52. Only if that's one of the options I guess. Good luck!

Volgvan3 years ago

6 0

Answer:

(25/4)∏ or (6 1/4)∏

Step-by-step explanation:

You might be interested in

A sample of n = 9 college students is used to evaluate the effectiveness of a new Study Skills Workshop. Each student’s grade

forsale [732]

Answer:

B) μD = 0.60 ± 0.10(1.397)

Step-by-step explanation:

The confidence interval is given by:

$MD±t_{\alpha/2, n-1} \frac{s}{\sqrt{n} }$

Where

MD=60

n=9

df=n-1=8

$t_{\alpha/2, n-1}=1.397$

$s=\sqrt{0.09} =0.3$

Then the confidence interval is

μD=0.60±1.397*(0.3/√9)

μD=0.60±0.10*(1.397)

6 0

3 years ago

Help

mars1129 [50]

To find one year, here's the equation:
5000 + 0.06(5000)
For 10 years:
5000 + 10(0.06(5000))
Multiply:
5000 + 0.6(5000)
We can make it smaller:
1.6(5000) = 8000
You can make $8000

7 0

3 years ago

Jim rented a truck for one day. There was a base fee of $18.95 , and there was an additional charge of 78 cents for each mile dr

erastovalidia [21]

18 - 15.95 = 2.05

2.05 divided by 71 =

2.9 miles

5 0

3 years ago

The population of a town increases at a rate of 5% every year. The population of this town was 10,400 in 2017. What will the pop

galben [10]

Answer:

The population of the town in 2020 will approx. be 12,039.

Step-by-step explanation:

The initial population in the year 2017 = 10,400

Increase in population every yea r = 5%

Now, 5% of 10, 400 = $\frac{5}{100} \times 10, 400 = 520$

Hence, the population in the year 2018 = 10,400 + 520 = 10,920

Now, in the year 2019

5% of 10, 920 = $\frac{5}{100} \times 10, 920 = 546$

Hence, the population in the year 2019 = 10,920 + 546 = 11,466

Now, in the year 2020

5% of 11,466 = $\frac{5}{100} \times 11,466 = 573.3$

or, the population in the year 2020 = 11,466 + 573.3 = 12,039.3

The population of the town in 2020 will approx. be 12,039.

6 0

3 years ago

The fish population of Lake Collins is decreasing at a rate of 5% per year. In 2004 there were about 1,350 fish. Write an expone

ad-work [718]

Answer:

Part 1) The exponential function is equal to $y=1,350(0.95)^{x}$

Part 2) The population in 2010 was $992\ fish$

Step-by-step explanation:

Part 1) Write an exponential decay function that models this situation

we know that

In this problem we have a exponential function of the form

$y=a(b)^{x}$

where

y ----> the fish population of Lake Collins since 2004

x ----> the time in years

a is the initial value

b is the base

we have

$a=1,350\ fish$

$b=(100\%-5\%)=95\%=0.95$

substitute

$y=1,350(0.95)^{x}$ ----> exponential function that represent this scenario

Part 2) Find the population in 2010

we have

$y=1,350(0.95)^{x}$

For $x=(2010-2004)=6\ years$

substitute

$y=1,350(0.95)^{6}=992\ fish$

7 0

3 years ago