Solve the equation -6a(-5-2a)=0

Write the standard equation of a circle that passes through (-5 5) with center (-10 -5) brainly

timama [110]

Answer:

The equation of the circle is (x + 10)² + (y + 5)² = 125 in standard form

Step-by-step explanation:

* lets study the standard form of the equation of a circle

- If the coordinates of the center of the circle are(h , k) and its radius

is r, then the standard equation of the circle is:

(x - h)² + (y - k)² = r²

* Now lets solve the problem

∵ The coordinates of the center of the circle are (-10 , -5)

∵ The standard form of the equation is (x - h)² + (y - k)² = r²

∵ h , k are the coordinates of the center

∴ h = -10 , k = -5

∴ The equation of the circle = (x - -10)² + (y - -5)² = r²

∴ The equation of the circle = (x + 10)² + (y + 5)² = r²

- To find the value of the radius lets use the point (-5 , 5) to

substitute their coordinate instead of x and y in the equation

∵ The circle passes through point (-5 , 5)

∵ (x + 10)² + (y + 5)² = r²

- Use x = -5 and y = 5

∴ (-5 + 10)² + (5 + 5)² = r² ⇒ simplify

∴ (5)² + (10)² = r²

∴ 25 + 100 = r²

∴ r² = 125

* Now lets write the equation in standard form

∴ (x + 10)² + (y + 5)² = 125

* The equation of the circle is (x + 10)² + (y + 5)² = 125 in standard form

8 0

3 years ago

A rectangle has vertices at (0, 0), (3, 0), and (0,6). What is the area of the rectangle?

Alexxandr [17]

First, set the coordinate points on a graph, and joint them to form a graph

By looking at the graph we can see that:

width : 3

length: 6

Area of a rectangle : Length x width : 6 x 3 = 18

7 0

1 year ago

A radio station had 60 tickets to a concert. They gave away 2 times as many tickets to listeners as to employees. How many ticke

Nataly_w [17]

Answer:

they gave 20 to the employee trust me

5 0

3 years ago

A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of

Ivenika [448]

The equation for the height of the rocket at time t given

$h= -16t^2+192t$

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

$h= -16t^2+192t$

$560 = -16t^2+192t$

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

$560 = -16(t^2 - 12t)$

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

$560/-16 = -16(t^2-12t)/-16$

$-35 = t^2 -12t$

Now we will move -35 to the righ side by adding 35 to both sides.

$-35+35 = t^2-12t+35$

$0 = t^2 -12t+35$

$t^2-12t+35 = 0$

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

$t^2-12t+35 =0$

$(t-5)(t-7) =0$

So by using zero product property we will get

$t-5 =0$

$t-5+5 = 0+5$

$t=5$

Also $t-7 =0$

$t-7+7 = 0+7$

$t=7$

So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

$h= -16t^2+192t$

$0= -16t^2 + 192 t$

$0 = -16(t^2-12t)$

$-16(t^2-12t) = 0$

We will move -16 to the other side by dividing it to both sides.

$-16(t^2-12t)/-16 = 0/-16$

$t^2-12t = 0$

We will take out the common factor t from the left side. By taking out t we will get,

$t(t-12) = 0$

We will use zero product property now. By using that we will get,

$t = 0$

ans also $t-12 = 0$

$t-12+12 = 0+12$

$t = 12$

When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.

6 0

3 years ago

HELP ME PLEASEEE!! 48 degrees 96 degrees x =

Roman55 [17]

Answer:

144°

96+48(further away angles)

8 0

3 years ago

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