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3 years ago

Write an equation for a circle with a radius of 3 and a center located at point (2,-1)

Mathematics

1 answer:

jeyben [28]3 years ago

4 0

Answer:

(x - 2)^2 + (y + 1)^2 = 9

Step-by-step explanation:

recall the equation for a circle:

(x - h)^2 + (y - k)^2 = r^2

here (h, k) = (2, -1)

and r = 3

(x - 2)^2 + (y -(-1))^2 = 3^2

(x - 2)^2 + (y + 1)^2 = 3^2

(x - 2)^2 + (y + 1)^2 = 9

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Harlene tosses two number cubes. If a sum of 8 or 12 comes up, she gets 9 points. If not, she loses 2 points. What is the expect

Amanda [17]

P(8 or 12) = 1/6
P(not 8 or 12 = 5/6
Let the expected points fro one roll be X.
$E(X)=9\times\frac{1}{6}-2\times\frac{5}{6}=-\frac{1}{6}\ points$
The answer is -(1/6) points.

5 0

2 years ago

Read 2 more answers

PLEASE HElp asap Solve for x. x = [?] X + 19 7x + 9

Bess [88]

X + 19 + 7x + 9 = 180
8x + 28 = 180
8x = 180 - 28
8x = 152
X = 152 : 8
X = 19

3 0

2 years ago

Maria kept a log of the number of miles she jogged each time she went for a run as she trained for an upcoming marathon. She dis

CaHeK987 [17]

Answer:

C. 1 to 7.99

Step-by-step explanation:

Let's recall that for calculating the range of a data set, we have to find out the lowest and the highest values of the data set.

As we can see in the graph below, the lowest value in the histogram is 1 and the highest is 7.99, therefore the correct answer is:

C. 1 to 7.99

4 0

2 years ago

Read 2 more answers

Consider the function represented by the equation 6q = 3s - 9. Write the equation in function notation, where q is the independe

Llana [10]

Q = 3s - 9/6
q = 3(s - 3)/6
q = s - 3/2

6q + 9 = 3s
6q + 9/3 = s
3(2q + 3)/3 = s
2q + 3 = s
s = 2q + 3

5 0

3 years ago

Let W be the subspace of R5 spanned by the vectors w1, w2, w3, w4, w5, where

Gwar [14]

To find W⊥, you can use the Gram-Schmidt process using the usual inner-product and the given 5 independent set of vectors.

Define projection of v on u as 
p(u,v)=u*(u.v)/(u.u) 
we need to proceed and determine u1...u5 as: 
u1=w1 
u2=w2-p(u1,w2) 
u3=w3-p(u1,w3)-p(u2,w3) 
u4=w4-p(u1,w4)-p(u2,w4)-p(u3,w4) 
u5=w5-p(u4,w5)-p(u2,w5)-p(u3,w5)-p(u4,w5) 

so that u1...u5 will be the new basis of an orthogonal set of inner space. 

However, the given set of vectors is not independent, since 
w1+w2=w3, 
therefore an orthogonal basis cannot be found.

7 0

3 years ago