All categories

3 years ago

4(x-3)=2(3x+1) 2 solve for x

1 answer:

3 0

Hi there!

Th' Eqn. is :-

4 (x - 3) = 2 (3x + 1)

=> 4x - 12 = 2 (3x + 1)

=> 4x - 12 = 6x + 2

Combine th' like terms :-

=> 4x - 6x = 2 + 12

=> - 2x = 14

=> x = $\dfrac {- 14}{2}$

=> x = - 7

Hence, The required answer is : x = - 7

~ Hope it helps!

You might be interested in

Please help ASAP ! Please help ASAP !

mihalych1998 [28]

Answer:

5/7

Step-by-step explanation:

4/7+1/7 is 5/7 and 5/7 is its simplest form. Hope this helps :)

5 0

3 years ago

Read 2 more answers

I need help to see if this is right

Marrrta [24]

The first thing I would like to help you with is recommend you buy a folder. It will help you keep your work organized and prevent catastrophes like this.

1
12.5 km/h
2
90 km/h
3
It would take about 4.5 seconds.
4
40 minutes
5
A. 0.2km/m
B. 0.5km/m
C. 0.3km/m faster

7 0

3 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

ANTONII [103]

Answer:

the answer is (3,4)

Step-by-step explanation:

because if you check wich quadrant it is in and you check the number you can see which one it is. x,y that is how it goes so 3,4 should be correct if you line them up.

6 0

2 years ago

Which of the following expressions are equivalent to 1/-7 x -6/5?

Aleks [24]

Answer:

Step-by-step explanation:

1/-7 × -6/5 = 6/35

A. -1/7 × 6/5 = -6/35

B. -6 × -7/5 = 42/5

Answer: C.

6 0

2 years ago

Read 2 more answers

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

$y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{o}$ - Initial height of the ball, measured in feet.

$v_{o}$ - Initial speed of the ball, measured in feet per second.

$g$ - Gravitational constant, equal to $-32.174\,\frac{ft}{s^{2}}$ .

$t$ - Time, measured in seconds.

Given that $y_{o} = 0\,ft$ , $v_{o} = 95\,\frac{ft}{s}$ , $g = -32.174\,\frac{ft}{s^{2}}$ and $y = 120\,ft$ , the following second-order polynomial is found:

$120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

$-16.087\cdot t^{2} + 95\cdot t -120 =0$

The roots of this polynomial are, respectively:

$t_{1} \approx 4.075\,s$ and $t_{2} \approx 1.831\,s$ .

Both roots solutions are physically reasonable, since $t_{1}$ represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas $t_{2}$ represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.

8 0

4 years ago