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Ainat [17]
3 years ago
14

Does anyone know the equation to this graph

Mathematics
2 answers:
faltersainse [42]3 years ago
4 0

Answer:

y = -x - 2

Step-by-step explanation:

You can start with the equation of a line in the slope-intercept form.

y = mx + b

m = slope

b = y-intercept

Look at the graph to find the y-intercept. At what point on the y-axis does the graph intersect the y-axis? Answer: -2. The y-intercept is -2. The value of b is -2.

Now we hjave

y = mx + (2)

or simply

y = mx - 2

Now we need to find the slope.

The slope is the ratio of the rise (difference in y) over the run (difference in x).

Look at the point (6, -7) that is shown in your graph. Go up 1 unit vertically. That is a rise of 1. Now go left 1 unit horizontally. That is a run of -1.

slope = m = rise/run = 1/(-1) = 1

The slope is 1. Now we replace m with -1.

y = mx - 2

y = -1x - 2

y = x - 2

Answer: y = -x - 2

Eduardwww [97]3 years ago
3 0

Answer: The equation of this graph can be written in slope intercept form which is what I am assuming you want.  If you do not want that form you can convert to any other form that you so desire.  The slope intercept equation for this graph would be y = -1x - 2.  The -1 is the slope of the line and the -2 is the y intercept.

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SashulF [63]

Answer:

168.5 m/s^2

Step-by-step explanation:

6 0
3 years ago
Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1
mafiozo [28]

Answer:

(a).   y'(1)=0  and    y'(2) = 3

(b).  $y'(t)=kb2t\cos(bt^2)$

(c).  $ b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

Step-by-step explanation:

(a). Let the curve is,

$y(t)=k \sin (bt^2)$

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value x_{0}  which lies in the domain of f where the derivative is 0.

Therefore,  y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function,    $y(t)=k \sin (bt^2)$

Differentiating the above equation with respect to x, we get

y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]

Applying chain rule,

y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)  

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

$y(t)=k \sin (bt^2)$

$y'(t)=kb2t\cos (bt^2)$

Now putting the initial conditions in the equation y'(1)=0

$kb2(1)\cos(b(1)^2)=0$

2kbcos(b) = 0

cos b = 0   (Since, k and b cannot be zero)

$b=\frac{\pi}{2}$

And

y'(2) = 3

$\therefore kb2(2)\cos [b(2)^2]=3$

$4kb\cos (4b)=3$

$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$

$2k\pi\cos 2 \pi=3$

2k\pi(1) = 3$  

$k=\frac{3}{2\pi}$

$\therefore b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

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4 years ago
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tester [92]

Answer:

Step-by-step explanation:

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4 years ago
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1 year ago
Please help with how I am supposed to do this. D:
kakasveta [241]

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