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2 years ago

The rate of change is constant in the graph Find the rate of change Explain what the rate of change means for the situation

Mathematics

2 answers:

Karolina [17]2 years ago

7 0

Answer:

The rate of of change is -100/3, or -33.33. This means that every year, the resale value of a refridgerator goes down by about $33.33

Step-by-step explanation:

rate of change formula : f(b)-f(a)/b-a

200-300/6-3

= -100/3

=-33.33

Westkost [7]2 years ago

5 0

The rate of change here, is the how much the value of the refridgerator increases per unit of time, in years. Since, the value of the fridge decreases, the rate of change is negative.

It is a linear graph i.e. the instantaneous rate of change and average rate of change are the same. Now, all you have to do is pick any two arbitrary points, preferably the x-intercept and y-intercept for simplicity and find the change in y and the change in x. The change in y and x will be the difference of the x and y coordinate of the points you chose. It does not matter which order you subtract in. Just remain consistent between your 2 points. The change in y is your rise, the change in x is your run. Slope is rise over run. So then divide.

Δy=400-0= $400

Δx = 0-12 = -12 yrs

Rate of change = rise/run = 400/(-12)=100/-3= -33.333 $/yrs

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2 years ago

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3 years ago

Read 2 more answers

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

2 years ago

solve the logarithmic equation using properties of logs. Round to the nearest thousandth if necessary log 2x+logx=2

Rudiy27

Yee
(apologies in advance for foematring. I'm on mobile)

log rules

if no base is stared assume base 10

log_a(b)=c means a^c=b

log(a)+log(b)=log(ab)

if a^b=c^b and b=b then a=c

so
log(2x)+log(x)=2
log_10(2x^2)=2
10^2=2x^2
100=2x^
50=x^2
$x = \pm 5\sqrt{2}$

6 0

2 years ago