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2 years ago

Anyone can help please and quickkkk

Mathematics

1 answer:

melomori [17]2 years ago

5 0

The values of x, y, and z are 1, 2, and 3 respectively after using the substitution method.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

We have a linear equation in three variable:

2x + 3y - z = 5 ...(1)

4x - y - z = -1 ...(2)

x + 4y + z = 12 ...(3)

$\rm \mathrm{Substitute\:}x=\dfrac{5-3y+z}{2}$ (from the equation 1 and 2)

After solving:

$\rm \begin{bmatrix}-7y+z+10=-1\\ \dfrac{5y+3z+5}{2}=12\end{bmatrix}$

From the above, two equations:
y =2

z =3

Plug the above values in equation 1

x = 1

Thus, the values of x, y, and z are 1, 2, and 3 respectively after using the substitution method.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ1

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2 years ago

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3 years ago

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3 years ago

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A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2.

IRISSAK [1]

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ( $v_0$ ) = 40 ft/sec

Deceleration of the car ( $\frac{dv}{dt}$ ) = -10 ft/sec²

Final speed of the car ( $v_x$ ) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, $\frac{dv}{dt}=-10\ ft/sec^2$

Negative sign means the velocity is decreasing with time.

Now, $\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})$ using chain rule of differentiation. Therefore,

$\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx$

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

$\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft$

Therefore, the car travels a distance of 80 feet before stopping.

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3 years ago