The average change in stocks value dropped $248 in the 5-day work week. What is the average change per day?

Write the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5).

lys-0071 [83]

Answer:

$(x+1)^2 + (y- 3.5)^2 =64$

Step-by-step explanation:

We are given that endpoints of a diameter at the points (7,2) and (-9,5).

Center of the circle is the mid point of diameter

So, first find the mid point of diameter

Formula : $x=\frac{x_1+x_2}{2} , y=\frac{y_1+y_2}{2}$

$(x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)$

Substitute the values in formula

$x=\frac{7-9}{2} , y=\frac{2+5}{2}$

$x=-1 , y=3.5$

So, center of circle = (h,k )=(-1,3.5)

To find length of diameter :

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)$

$d=\sqrt{(-9-7)^2+(5-2)^2}$

$d=\sqrt{265}$

Length of radius = r = $\frac{Diameter}{2}=\frac{\sqrt{256}}{2}$

Standard form of the equation of the circle : $(x -h)^2 + (y- k)^2 = r^2$

(h,k )=(-1,3.5)

$r=\frac{\sqrt{256}}{2}$

Equation of the circle : $(x+1)^2 + (y- 3.5)^2 =(\frac{\sqrt{256}}{2})^2$

Equation of the circle : $(x+1)^2 + (y- 3.5)^2 =64$

Hence the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5) is $(x+1)^2 + (y- 3.5)^2 =64$

5 0

3 years ago

Can someone help me with this?

Elden [556K]

Good evening ,

Answer:

{(1,6),(2,2),(4,7),(3,2)}

Step-by-step explanation:

As a function , f has a unique image for each number of the domain.

for instance, the first set can’t represent a function because 1 has more than one image(number).

:)

7 0

3 years ago

3z-4=6z-17 how do solve

Maru [420]

3z-4=6z-17
-3z-4=-17 (subtracted the 6z from both sides)
-3z=-13 (added the 4 to both sides)
z= $\frac{-13}{-3z}$ (divided both sides by -3z)
z= $\frac{13}{3}$

5 0

3 years ago

Read 2 more answers

Pleaseee helppp with thissss asapppp

Natasha2012 [34]

As we know that the standard equation of circle is ${\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}}$ , where r is the radius of circle and centre at (h,k) 

Now , as the circle passes through (2,9) so it must satisfy the above equation after putting the values of h and k respectively

${:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}$