Plz help me with this

8. The straight line depreciation equation for a car is

kifflom [539]

An algebraic expression to represent the value of this car after M months is y=Mx+C.

We have given that the

The straight-line depreciation equation for a car is,

y = -2,680x + 26,800.

therefore after 4moths

y = -2,680(4) + 26,800.

y = -10720 + 26,800.

y=16080

<h3>What is the slope?</h3>

The slope of a line is a measure of its steepness. Mathematically, the slope is calculated as rise over run (change in y divided by change in x)

b. For find after the 75 months replace x by 75 and slolve the given inequality.

Suppose that M represents the length of time in months when

the car still has value.

An algebraic expression to represent the value of this car after M months.

y=Mx+C.

To learn more about the line visit:

brainly.com/question/1821791

#SPJ1

5 0

2 years ago

Jennifer pours 1 over 2 quart of milk equally into 4 glasses. How much milk, in quarts, does Jennifer pour into each glass?

Kitty [74]

Answer:

1/8 quart

Step-by-step explanation:

<u>Volume of milk:</u>

1/2 quart

<u>Milk in each glass:</u>

1/2 : 4 = 1/8 quart

1/8 quart of milk into each glass

4 0

4 years ago

Pls pls pls pls pls pls pls pls pls pls pls

Nikitich [7]

I believe it would be 130 degrees

5 0

3 years ago

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If w = 5 cos (xy) − sin (xz) and x = 1/t , y = t, z = t^3 ; then find dw/dt

Scrat [10]

In this question, we find the derivatives, using the chain's rule.

Doing this, the derivative is:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Chain Rule:

Suppose we have a function $w(x,y,z), x = x(t), y = y(t), z = z(t)$ , and want to find it's derivative as function of t. It will be given by:

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

Thus, we have to find the desired derivatives, which are:

w of x:

$\frac{dw}{dx} = -5y\sin{(xy)} - z\cos{(xz)}$

Considering $x = \frac{1}{t}, y = t, z = t^3$

$\frac{dw}{dx} = -5t\sin{(1)} - t^3\cos{(t^2)}$

w of y:

$\frac{dw}{dy} = -5x\cos{(xy)}$

Considering $x = \frac{1}{t}, y = t$

$\frac{dw}{dy} = -\frac{5}{t}\cos{1}$

w of z:

$\frac{dw}{dz} = -x\cos{(xz)}$

Considering $x = \frac{1}{t}, z = t^3$

$\frac{dw}{dz} = -\frac{1}{t}\cos{(t^2)}$

Derivatives of x, y and z as functions of t:

$\frac{dx}{dt} = -\frac{1}{t^2}$

$\frac{dy}{dt} = 1$

$\frac{dz}{dt} = 3t^2$

Derivative of w as function of t.

Now, we just replace what we found into the formula. So

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

$\frac{dw}{dt} = (-5t\sin{(1)} - t^3\cos{(t^2)})(-\frac{1}{t^2}) - (\frac{5}{t}\cos{1}) - (\frac{1}{t}\cos{(t^2)})3t^2$

Applying the multiplications:

$\frac{dw}{dt} = \frac{5}{t}\sin{1} + t\cos{t^2} - \frac{5}{t}\cos{1} - 3t\cos{t^2}$

Applying the simplifications:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Which is the derivative.

For more on the chain rule, you can check brainly.com/question/12795383

8 0

3 years ago

−4x + 15= -13<br> what is x?

skelet666 [1.2K]

$-4x+15=-13 \ \ \ |-15 \\ \\ -4x=-28 \ \ \ |:(-4) \\ \\ x=7$

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4 0

3 years ago

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