All categories

3 years ago

What is the center of the circle that has a diameter whose endpoints are (6, -4) and (18, 10)?

Mathematics

2 answers:

Maurinko [17]3 years ago

8 0

Answer:

(12, 3)

Step-by-step explanation:

To answer this, we find the midpoint of the line segment (that is, of the diameter) connecting (6, -4) and (18, 10):

The formula for the midpoint gives us x-midpoint = (6+18)/2, or x-midpoint = 12.

The formula for the midpoint gives us y-midpoint = (-4+10)/2, or x-midpoint = (6)/2 = 3.

Thus, the center of this circle is at (x-midpoint), y-midpoint), or (12, 3).

The third answer choice is the correct one.

Leno4ka [110]3 years ago

7 0

Answer:

(12, 3)

Step-by-step explanation:

You might be interested in

Charlotte is on spring break in Mexico. She purchases a souvenir for 153 pesos, which she charges to her credit card. If the exc

KATRIN_1 [288]

1.) multiply the US dollars by the exchange rate $213 x 80.76 = 17,201.88

2. )divide pesos by the exchange rate 158/11.69 = $13.5158 so (a) is the correct answer

So the answer is

$13.52

6 0

3 years ago

A graph is shown below <br> what is the equation of the line in slope intercept form HELP ASAP

JulijaS [17]

Answer:

y = -2x + 16

the y intercept is the y value at x = 0. On the graph there is a dot at 16 on x= 0, so the y intercept would be 16.

the only answer choice adding 16 sis y = -2x + 16

7 0

2 years ago

Can someone help me with this

arsen [322]

Answer:

1. 40 members

2. 4 cheerleaders

3. 40 marbles

4. 2 marbles

5. 8

6. 24

Step-by-step explanation:

1. cross multiply 12/x=30/100

2. cross multiply 12/x=10/100

3. 18/x=45/100

4. x/40=5/100

5. we know 5% is 2 marbles. multiply by 4 to get 8 marbles.

6. 20%=8 marbles. 60/20= 3. 3 x 8= 24 marbles.

6 0

2 years ago

Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

melamori03 [73]

Answer:

The complete solution is

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

$y = e^{mx}$

Differentiating with respect to x

$y'= me^{mx}$

Again differentiating with respect to x

$y''= m ^2 e^{mx}$

Putting the value of y, y' and y'' in left side of the differential equation

$3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0$

$\Rightarrow 3m^2-8m-3=0$

The auxiliary equation is

$3m^2-8m-3=0$

$\Rightarrow 3m^2 -9m+m-3m=0$

$\Rightarrow 3m(m-3)+1(m-3)=0$

$\Rightarrow (3m+1)(m-3)=0$

$\Rightarrow m = 3, -\frac13$

The complementary function is

$y= Ae^{3x}+Be^{-\frac13 x}$

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)

For particular integral

$y_p=\frac 1{(3D+1)(D-3)} .4$

$=4.\frac 1{(3D+1)(D-3)} .e^{0.x}$ [since $e^{0.x}=1$ ]

$=4\frac{1}{(3.0+1)(0-3)}$ [ replace D by 0 , since L(0)≠0]

$=-\frac43$

The complete solution is

y= C.F+P.I

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

4 0

3 years ago

What inequality is on the number line?

Elis [28]

Answer:
y ≥ -5

Explanation:
y is greater than -5 because the darkened line is heading towards the positive side.
(The positives are greater than the negatives.)
(The smaller negative numbers are greater than the larger negative numbers.)

7 0

2 years ago