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2 years ago

Find the equation of the circle with center at the origin that contains the point (-8,6)

1 answer:

3 0

The center is at origin O(0,0).
If it contains the point, P(-8,6), then the radius r is, by Pythagoras theorem,
r=sqrt((-8)^2+6^2)=10

The general equation of a circle at centre (xc,yc) with radius r is given by
(x-xc)^2+(y-yc)^2=r^2

Substituting r=10, (xc,yc)=(0,0)
the resulting equation is therefore

(x-0)^2+(y-0)^2=10^2
or simply
x^2+y^2=100

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What is the rule of a function of the form f(t)= a sin (bt+c) +d whose graph appears to be identical to the given graph?

Gnoma [55]

Answer:

Option D.

Step-by-step explanation:

We can easily solve this problem by using a graphing calculator or plotting tool.

The function is

f(t) = a*sin (b*t +c) + d

Please, see attached picture below.

By looking at the picture with all the possible cases, we can tell that the correct option is D.

The function has a period of T = 6π

Max . Amplitude = 9

Min . Amplitude = -5

7 0

3 years ago

Read 2 more answers

How many times does 5/8 go into 3 3/4?

muminat

5/8 times what=3 and 3/4
convert 3 and 3/4 to improper
3 and 3/4=3+3/4=12/4+3/4=15/4

5/8 times what=15/4
times both sides by 8/5
what=120/20
what=12/2
what=6

6 times

8 0

3 years ago

padilas [110]

Answer: B

Trust me I took the test right now

6 0

2 years ago

1/8 + c = 4/5 <br><br>what is c

ASHA 777 [7]

Answer:

27/40

Step-by-step explanation:

7 0

2 years ago

For this exercise assume that all matrices are ntimesn. Each part of this exercise is an implication of the form "If "statement

inna [77]

Answer:

C. True; by the Invertible Matrix Theorem if the equation Ax=0 has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the n x n identity matrix.

Step-by-step explanation:

The Invertible matrix Theorem is a Theorem which gives a list of equivalent conditions for an n X n matrix to have an inverse. For the sake of this question, we would look at only the conditions needed to answer the question.

There is an n×n matrix C such that CA= $I_n$ .
There is an n×n matrix D such that AD= $I_n$ .
The equation Ax=0 has only the trivial solution x=0.
A is row-equivalent to the n×n identity matrix $I_n$ .
For each column vector b in $R^n$ , the equation Ax=b has a unique solution.
The columns of A span $R^n$ .

Therefore the statement:

If there is an n X n matrix D such that AD=I, then there is also an n X n matrix C such that CA = I is true by the conditions for invertibility of matrix:

The equation Ax=0 has only the trivial solution x=0.

A is row-equivalent to the n×n identity matrix $I_n$ .

The correct option is C.

5 0

3 years ago