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3 years ago

What is a correct method of solving the equation x/3-6=9 for x? please help thx

Mathematics

1 answer:

stira [4]3 years ago

5 0

Answer:

Step-by-step explanation:

$\frac{x}{3} -6=9$

add 6 on both sides to eliminate it(doing this will get us closer to the variable)

x/3=15

multiply both sides by 3

x=45

hope this helps!

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If a line crosses the y-axis at (0, 1) and has a slope of 45, what is the equation of the line?

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The equation for a linear function is y = mx + b, where m is the slope and b is the y-intercept.
The equation would be y = 45x + 1.

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3 years ago

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maksim [4K]

The answer for this equation is q=5

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3 years ago

a water cooler shaped like a rectangular prism has length of 80 centimeters a width of 40 centimeters and height of 100 centimet

maks197457 [2]

$volume = lwh$

$volume = (80)(40)(100)$

$volume = 320000 \: {cm}^{3}$

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8 0

3 years ago

A car dealership decreased the price of a certain car by 4% . The original price was $43,600 . write the new price in terms of t

defon

Answer: The new price of the car is $41856

Step-by-step explanation:

So we know the the original price as 43,600 which is 100% and is being dropped by 4% so you would have to subtract 4% from a 100% and multiply it by the original price.

100% - 4% = 96%

Now 96% of the original price is the new price.

96% * 43,600= ?

0.96 * 43,600 = 41856

6 0

3 years ago

In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates

Anettt [7]

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

$S = ut$ + $\frac{1}{2} at^{2}$

Where $S$ is distance

$u$ is the initial velocity

$a$ is acceleration

and $t$ is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

$S$ = S₁

$u$ = 0 m/s ( The player starts at rest)

$a$ = 2.2 m/s²

Then,

S₁ = $0(t)$ + $\frac{1}{2} (2.2)t^{2}$

S₁ = $1.1t^{2}$

$t^{2} = \frac{S_{1} }{1.1}$

For player 2

$S$ = S₂

$u$ = 0 m/s ( The player starts at rest)

$a$ = 1.7 m/s²

Then,

S₂ = $0(t)$ + $\frac{1}{2} (1.7)t^{2}$

S₂ = $0.85t^{2}$

$t^{2} = \frac{S_{2} }{0.85}$

Since the time spent by both players is equal, We can write that

$t^{2}$ = $t^{2}$

Then,

$\frac{S_{1} }{1.1} = \frac{S_{2} }{0.85}$

$0.85S_{1} = 1.1S_{2}$

From, S₁ + S₂ = 41 m

S₂ = 41 - S₁

Then,

$0.85S_{1} = 1.1 ( 41 -S_{1} )$

$0.85S_{1} = 45.1 - 1.1S_{1}$

$0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1} = 23.13 m$

This is the distance covered by the first player. We can then put this value into $t^{2} = \frac{S_{1} }{1.1}$ to determine how much time elapses before the players collide.

$t^{2} = \frac{S_{1} }{1.1}$

$t^{2} = \frac{23.13 }{1.1}$

$t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs$

Hence, the time that elapses before the players collide is 4.59 secs

3 0

4 years ago