Ask question

Ask question

All categories

1 year ago

10

Need to solve this !!

1 answer:

ValentinkaMS [17]1 year ago

6 0

We just went over this in class and it’s actually quite easy once you get your head around it!

Im not quite sure what the second one it asking but here’s an idea :)

you need them all to be on the same side so you can do
x^2 - 4x - 12 = 0
(x-6)(x+2) = 0
x = 6 or 2
(each bracket must equal 0)

You might be interested in

What is the standard form that has a center (1,3) and a radius of 9

klasskru [66]

$\huge\boxed{(x-1)^2+(y-3)^2=9^2}$

Start with the standard form of a circle, where $(h,k)$ is the center of the circle and $r$ is the radius.

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

Plug in the known values.

$\boxed{(x-1)^2+(y-3)^2=9^2}$

This is the circle's equation in standard form.

4 0

2 years ago

Help and tell me why

lorasvet [3.4K]

Answer:

60

Step-by-step explanation:

V= lwh / 3 Write formula

V= 6*6*5 /3 Substitute

V=60

3 0

2 years ago

I am trying to figure out how to simplify this problem .

PIT_PIT [208]

Answer:

Step-by-step explanation:11+12/square root of 16

11+(12/4)

11+3

answer is 14

7 0

1 year ago

Read 2 more answers

One more question on this test guys. please answer as fast as possible. If you do,

GaryK [48]

1. The given rectangular equation is $x=2$ .

We substitute $x=r\cos \theta$ .

$r\cos \theta=2$

Divide through by $\cos \theta$

$r=\frac{2}{\cos \theta}$

$r=2}\sec \theta$

$\boxed{x=2\to r=2\sec \theta}$

2. The given rectangular equation is:

$x^2+y^2=36$

This is the same as:

$x^2+y^2=6^2$

We use the relation $r^2=x^2+y^2$

This implies that:

$r^2=6^2$

$\therefore r=6$

$\boxed{x^2+y^2=36\to r=6}$

3. The given rectangular equation is:

$x^2+y^2=2y$

This is the same as:

We use the relation $r^2=x^2+y^2$ and $y=r\sin \theta$

This implies that:

$r^2=2r\sin \theta$

Divide through by r

$r=2\sin \theta$

$\boxed{x^2+y^2=2y\to r=2\sin \theta}$

4. We have $x=\sqrt{3}y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta\sqrt{3}$

This implies that;

$\tan \theta=\frac{\sqrt{3}}{3}$

$\theta=\frac{\pi}{6}$

$\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}$

5. We have $x=y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta$

This implies that;

$\tan \theta=1$

$\theta=\frac{\pi}{4}$

$\boxed{x=y\to \theta=\frac{\pi}{4}}$

6 0

2 years ago

A) The perimeter of a rectangle is the sum of the lengths of its four sides. Write an expression for the perimeter of the rectan

goblinko [34]

Answer:

Below

Step-by-step explanation:

The length of this triangle is 3x+1 and the width is x.

The perimeter P is:

P= 2(3x+1)+2*x

P= 6x+2+2x

P= 8x+2

Let's evaluate it when x=1/2

●1/2 =0.5

P= 8*0.5+2 =4+2= 6 ft

●●●●●●●●●●●●●●●●●●●●●●●●

The area A is:

A = (3x+1)*x

A= 3x^2 +x

Let's evaluate it when x=0.5 feet

A= 3*0.5^2 +0.5

A= 3*0.25+0.5

A= 0.75 +0.5

A= 1.25 ft^2

6 0

2 years ago