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3 years ago

( − 2 ( 4 ) x − 2 ) < − 8 x + 4

Mathematics

1 answer:

anastassius [24]3 years ago

7 0

Answer:

1. Expand.

8x-6-8=4+2x8x−6−8=4+2x

2. Simplify 8x-6-88x−6−8 to 8x-148x−14.

8x-14=4+2x8x−14=4+2x

3. Add 1414 to both sides.

8x=4+2x+148x=4+2x+14

4. Simplify 4+2x+144+2x+14 to 2x+182x+18.

8x=2x+188x=2x+18

5. Subtract 2x2x from both sides.

8x-2x=188x−2x=18

6. Simplify 8x-2x8x−2x to 6x6x.

6x=186x=18

7.Divide both sides by 66.

x=\frac{18}{6}x= 18/6

8. Simplify \frac{18}{6}

18 to 33.

x=3x=3

Step-by-step explanation:

You might be interested in

Use Newton's Method to approximate the zero(s) of the function. Continue the iterations until two successive approximations diff

liubo4ka [24]

Answer:

There is only one real zero and it is located at x = 1.359

Step-by-step explanation:

After the 4th iteration the solution was repeating the first 3 decimal places. The formula for Newton's Method is

$x_{n}-\frac{f(x_{n}) }{f'(x_{n}) }$

If our function is

$f(x)=x^5+x-6$

then the first derivative is

$f'(x)=5x^4+1$

I graphed this on my calculator to see where the zero(s) looked like they might be, and saw there was only one real one, somewhere between 1 and 2. I started with my first guess being x = 1.

When I plugged in a 1 for x, I got a zero of 5/3.

Plugging in 5/3 and completing the process again gave me 997/687

Plugging in 997/687 and completing the process again gave me 1.36976

Plugging in 1.36976 and completing the process again gave me 1.359454

Plugging in 1.359454 and completing the process again gave me 1.359304

Since we are looking for accuracy to 3 decimal places, there was no need to go further.

Checking the zeros on the calculator graphing program gave me a zero of 1.3593041 which is exactly the same as my 5th iteration!

Newton's Method is absolutely amazing!!!

5 0

3 years ago

Two airplanes left the same airport and arrived at the same destination at the same time. The first airplane left at 8:00 a.m. a

ddd [48]

The formula for distance is equal to:

d = v * t

where d is distance, v is velocity or speed, and t is time

Since the distance travelled by the two airplane is similar, therefore we can create the initial equation:

v1 * t1 = v2 * t2

We know that v1 = 496, and v2 = 558 so:

496 t1 = 558 t2

x = 558 t2 / 496

We also know that airplane 1 travelled 30 minutes (0.5 hours) earlier than airplane 2, therefore:

x = t2 + 0.5

Hence,

496 (t2 + 0.5) = 558 t2

496 t2 + 248 = 558 t2

t2 = 4 hours

x = t2 + 0.5 = 4 + 0.5

x = 4.5 hours

So the equation is:

x = 558 t2 / 496

And the first plane travelled:

x = 4.5 hours

5 0

3 years ago