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3 years ago

What is the center of the circle shown below?

Mathematics

2 answers:

Diano4ka-milaya [45]3 years ago

7 0

Answer:

A. S

Step-by-step explanation:

The edge of the circle would be M and there is a line to the center showing the radius. The second letter shows the center.

Kipish [7]3 years ago

7 0

Answer: S

Step-by-step explanation:

You might be interested in

Anyone know the answer to this?!?!

Ad libitum [116K]

Answer:

C: 3 x 4 x 4

Step-by-step explanation:

It's obvious for an European since it's simple approximation, the cm is just thrown in to confuse Americans.

6 0

3 years ago

Read 2 more answers

Why do you think items that come in a larger quantity cost less per unit

MAXImum [283]

They cost less because there are more items in stock meaning that there isn't a limit here is another way of thinking if there are less Items then they can charge you more per item because they know you are gonna want it but if there are tons of that one item they there kinda like eh then you think oh I can live without that....... Hope this helps

7 0

2 years ago

Read 2 more answers

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago

Seven less than the product of a number and 10 is 50

Veseljchak [2.6K]

product of a number and 10: 10n
Seven less than the product of a number and 10: 10n - 7
so
Seven less than the product of a number and 10 is 50:
10n - 7 = 50
----------------------------
if you need to find the number n then
10n - 7 = 50
10n = 57
n = 57/10
n = 5.7

5 0

3 years ago

find all integers x satisfying each of the following congruences mod n. if no such x exists, explain why not

zalisa [80]

Answer:

please find the solution:

Step-by-step explanation:

please find the correct question:

Given value:

$\to 4x \equiv 2(mod\ n) \ \ \ \ \ \ \ \ \ \ \ \ _{where} \ \ n=6\\\\\to 4x \equiv 2 (mod 6)$

$\to 4x-2$ should be divisible by 6 we are also given that $0 \leq x < n$ and start with x = 0

$\to 4x-2$ =-2 which is not divisible by 6

x= 1

$\to 4x-2$ = 2 not divisilble by 6

x= 2

$\to 4x-2$ = 6 which is divisible by 6

x=3

$\to 4x-2$ = 10 not divisible by 6

x=4

$\to 4x-2$ = 14 not divisible by 6

x=5

$\to 4x-2$ = 18 which is divisible by 6

so such x are x = 2 and x = 5

4 0

3 years ago