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gladu [14]
3 years ago
14

If the center of a circle is (1, 1), and it has a radius of 5, which is another point on the circle?

Mathematics
1 answer:
EleoNora [17]3 years ago
5 0
If this question is not multiple choice, another point on the circle could be (1,6) because it is 5 units away from (1,1).
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–4r − 10r + –5 = –19
defon

Answer:

r=1

Step-by-step explanation:

-4 r -10r + -5= -19

-4r + (-10r) + -5=-19

-14r+ -5 = -19

-14r + -5 +5= -19+5

-14r= -14

r=1

5 0
3 years ago
What is an explicit formula for the geometric sequence 64,16,4,1,... where the first term
olya-2409 [2.1K]

Answer:

A

Step-by-step explanation:

The explicit formula for a geometric sequence is

f(n) = f(1) (r)^{n-1}

where f(1) is the first term and r the common ratio

Here f(1) = 64 and r = \frac{f(2)}{f(1)} = \frac{16}{64} = \frac{1}{4} , then

f(n) = 64 (\frac{1}{4}) ^{n-1} → A

7 0
2 years ago
PLEASE HELP ASAP 30 PTS ):
ASHA 777 [7]

Answer:


Step-by-step explanation:

It's an exponential function but it is going in the negative direction. So now you know there has to be a negative in front of the function meaning it can only be c or d. Since we know its a exponential function. If we set x = 0, it will give us the y intercept. Which should be 1. However it is -2. Therefore it it multiplied by -2 meaning d is right answer

6 0
3 years ago
Read 2 more answers
Consider the initial value problem y′+2y=4t,y(0)=8.
Xelga [282]

Answer:

Please read the complete procedure below:

Step-by-step explanation:

You have the following initial value problem:

y'+2y=4t\\\\y(0)=8

a) The algebraic equation obtain by using the Laplace transform is:

L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\

next, you replace (1) and (2):

sY(s)-y(0)+2Y(s)=\frac{4}{s^2}\\\\sY(s)+2Y(s)-8=\frac{4}{s^2}  (this is the algebraic equation)

b)

sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)} (this is the solution for Y(s))

c)

y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}

To find the inverse Laplace transform of the first term you use partial fractions:

\frac{4}{s^2(s+2)}=\frac{-s+2}{s^2}+\frac{1}{s+2}\\\\=(\frac{-1}{s}+\frac{2}{s^2})+\frac{1}{s+2}

Thus, you have:

y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}  

(this is the solution to the differential equation)

5 0
3 years ago
Which of the following are continuous?
goldenfox [79]
E and D are continuous
7 0
3 years ago
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