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Tresset [83]
3 years ago
11

Which equation models a line that passes through point (-2,-2) and has a slope of -8

Mathematics
1 answer:
Nitella [24]3 years ago
5 0
The answer would be y=-8x-18
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Can you help me if u want
Ganezh [65]

Answer:

the correct symbol is: Δ

Step-by-step explanation:

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2 years ago
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What is the least common denominator of 1/7,2/5 and 2/3?
Maurinko [17]

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105 is the least common denominator

Step-by-step explanation:

Since 7, 5 and 3 are all prime.

LCM would be 7×3×5 = 105

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3 years ago
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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
Art [367]

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

6 0
3 years ago
Match each graph to the equation of its line.
Serjik [45]

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b

Step-by-step explanation:

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Uhh I need help what is equivalent to 1 to 4 ????​
Nataly_w [17]

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2/8

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