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2 years ago

Which function family is represented by the table? x -5 0 5 10 15 20 y 3 4 5 6 7 8

Mathematics

1 answer:

Ksenya-84 [330]2 years ago

6 0

Domain ( Inputs ) Range (Outputs)

2, 5, 10. 6, 15 , 30

X Y

2. 6

5. 15

10. 30

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Pls help i will give brainliest

Dominik [7]

Well first of all the shorter pipe would be 160ft bc 120/3 is 40 and 40*4 is 160 so the shorter piece is 160ft and 160+12 is 172 so the longer piece of pipe would be 172ft

I’m pretty sure I’m correct, lemme know if I’m not
Hope this helps :)

6 0

3 years ago

PLEASE HELP HELPPPPPPPPP

professor190 [17]

Answer:

x+2y=- 7

x-2y =9

Step-by-step explanation:

Solving Steps

2x+4y=-14

x-2y=9

Divide both sides

(2x+4y) ÷2=-14=2

{x-2y=9

Remove the parentheses

Divide

2x÷2+4y÷2=-7

x-2y =9

Divide

Solution

x+2y=- 7

x-2y =9

5 0

3 years ago

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21 by 36 subtract 2 by 5 multiple 5 by 8

Nikolay [14]

Answer:

<h2>That's my opinion </h2>

Step-by-step explanation:

<h3><em><u>Imsureeeeeeee123</u></em><em><u> </u></em></h3>

7 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

In Triangle uvw , the measure of

cricket20 [7]

Answer:

not enough info

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers