How can you write a fraction as a decimal?

X - (-6) =8<br> Please answer

sergeinik [125]

Answer:

X=2

Step-by-step explanation:

6 0

3 years ago

Help and please explain!

UkoKoshka [18]

Answer:

a. 146

Step-by-step explanation:

{4*[72-6*(3+2)]}-22

=[4*(72-6*5)]-22

=[4*(72-30)]-22

=(4*42)-22

=168-22

=146

4 0

3 years ago

Which equations are true for x = –2 and x = 2? Select two options x2 – 4 = 0 x2 = –4 3x2 + 12 = 0 4x2 = 16 2(x – 2)2 = 0

Svetllana [295]

The equation $x^{2} -4=0~~and~~4x^2=16$ are true for x = -2 and x = 2.

The given equations are given as:

$x^2-4=0\\\\x^2=-4\\\\3x^2+12=0\\\\4x^2=16\\\\2(x-2)^2=0$

We need to select two equations that are true for x = -2 and x = 2.

<h3>What are the solutions to an equation ?</h3>

The solutions of an equation are the values that satisfy the given equation or make the equation true when substituted for unknowns in the equations.

Example:

x - 2 = 2

Here the solution for x - 2 = 2 is 4 because x = 4 will make the equation true.

4 - 2 = 2

2 = 2

Let's find the solutions for each equation.

1.

$x^2 - 4=0$

$x^{2}$ = 4

x = $\sqrt{4}$ = ± 2

x = = 2 and x = -2

2.

$x^2 =-4$

x = $\sqrt{-4}$

x = $\sqrt{-1 \times 4}$ = $\sqrt{-1}\times\sqrt{4}$

x = ± 2 $\sqrt{-1}$

x = 2i and x = -2i where i = $\sqrt{-1}$

3.

3 $x^{2}$ + 12 = 0

$x^{2}$ = -12 / 3 = -4

x = $\sqrt{-1}$ $\sqrt{4}$

x = ± 2 $\sqrt{-1}$

x = 2i and x = -2i

4.

4 $x^{2}$ = 16

$x^{2}$ = 16 / 4

$x^{2}$ = 4

x = $\sqrt{4}$

x = ±2

x = 2 and x = -2

5.

$2(x-2)^2 = 0$

$(x-2)^2 = 0$

x - 2 = 0

x = 2.

Thus the equation $x^{2} -4=0~~and~~4x^2=16$ have solutions x = -2 and x = 2.

Learn more about solutions of equations here:

brainly.com/question/14506845

#SPJ1

4 0

2 years ago

пряма С перетинає пряму А і не перетинає пряму В паралельно прямій А яке взаємне розміщення прямих В і С

8090 [49]

Answer:

I dont even undertsand what you jst types

Step-by-step explanation:

qk fhflk

7 0

3 years ago

2 cos x +3 sin 2x = 0<br> answer in degrees

enot [183]

Answer:

If want just the approximated solutions in the interval from 0 to 360:

199.47

340.53

90

270

If you want all the approximated solutions:

199.47+360k

340.53+360k

90+360k

270+360k

Step-by-step explanation:

2 cos(x)+3 sin(2x)=0

First step: Use double angle identity for sin(2x). That is, use, sin(2x)=2sin(x)cos(x).

2 cos(x)+3*2sin(x)cos(x)=0

2 cos(x)+ 6sin(x)cos(x)=0

Factor the 2cos(x) out, like so:

2cos(x)[ 1 + 3 sin(x)]=0

In order for this product to be zero, we must find when both factors are 0.

2cos(x)=0 or 1+3sin(x)=0

Let's do 2cos(x)=0 first.

2cos(x)=0

Divide both sides by 2:

cos(x)=0

So the x-coordinate is 0 on the unit at x=90 deg and x=270 deg (in the first rotation).

Let's do 1+3sin(x)=0.

1+3sin(x)=0

Subtract 1 on both sides:

3sin(x)=-1

Divide both sides by 3:

sin(x)=-1/3

Unfortunately this is not on the unit circle so I'm just going to take sin^-1 or arsin on both sides (this is the same thing sin^-1 or arsin).

x=arcsin(-1/3)=-19.47 degrees

So that means -(-19.47)+180 is also a solution so 19.47+180=199.47 .

And that 360+-19.47 is another so 360+-19.47=340.53 .

So the solutions for [0,360] are

199.47

340.53

90

270

If you want all the solutions just add +360*k to each line where k is an integer.

4 0

3 years ago