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garik1379 [7]
3 years ago
12

Can someone check if this is correct 9th grade math algebra 1

Mathematics
1 answer:
vovangra [49]3 years ago
3 0

Answer:

I would but I'm no 9th grader my brother probably could tho...but he won't

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For the function in the graph, find the values of f(-4), f(-1), and f(1).​
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Answer:

f(-4) = -1

f(-1) = -4

f(1) = 4

Step-by-step explanation:

We need to find the y value for the x values

f(-4) means find the y value for x= -4

f(-4) = -1

f(-1) = -4

f(1) = 4

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1

Step-by-step explanation:

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"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e
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Answer:

a. v(t)= -6.78e^{-16.33t} + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=e^{\int\limits^  {}k \, dt } =e^{kt}. We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ e^{kt}v' + ke^{kt}v = ge^{kt} ⇒ [ve^{kt}]' = ge^{kt}. Integrating, we have

∫ [ve^{kt}]' = ∫ge^{kt}

    ve^{kt} = \frac{g}{k}e^{kt} + c

    v(t)=   \frac{g}{k} + ce^{-kt}.

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + ce^{-16.33 * 0} = 16.33 + c

       c = 9.55 -16.33 = -6.78.

So, v(t)=   16.33 - 6.78e^{-16.33t}. m/s = - 6.78e^{-16.33t} + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78e^{-16.33 x 0.5} + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

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