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lozanna [386]
3 years ago
8

How does graphing linear inequalities differ from graphing linear equations?​

Mathematics
1 answer:
Ivan3 years ago
6 0

Explanation:

When the inequality symbol is replaced by an equal sign, the resulting linear equation is the boundary of the solution space of the inequality. Whether that boundary is included in the solution region or not depends on the inequality symbol.

The boundary line is included if the symbol includes the "or equal to" condition (≤ or ≥). An included boundary line is graphed as a solid line.

When the inequality symbol does not include the "or equal to" condition (< or >), the boundary line is not included in the solution space, and it is graphed as a dashed line.

Once the boundary line is graphed, the half-plane that makes up the solution space is shaded. The shaded half-plane will be to the right or above the boundary line if the inequality can be structured to be of one of these forms:

  • x > ...   or   x ≥ ...   ⇒ shading is to the right of the boundary
  • y > ...   or   y ≥ ...   ⇒ shading is above the boundary

Otherwise, the shaded solution space will be below or to the left of the boundary line.

_____

Just as a system of linear equations may have no solution, so that may be the case for inequalities. If the boundary lines are parallel and the solution spaces do not overlap, then there is no solution.

_____

The attached graph shows an example of graphed inequalities. The solutions for this system are in the doubly-shaded area to the left of the point where the lines intersect. We have purposely shown both kinds of inequalities (one "or equal to" and one not) with shading both above and below the boundary lines.

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Nesterboy [21]

Answer: It is because if ∠N and ∠K are congruent and are same in size and shape.

8 0
2 years ago
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1) What is 20% of 20?<br> 2) what is 15 is what percent of 30?<br> 3) 50 is 50% of what?
N76 [4]

Hi there!

The word "of" is the same thing as a multiplication sign.

1)

20% = 0.2

0.2 × 20 = 4


2)

? × 30 = 15

divide each side of the equation by 30

? = 0.5

0.5 = 50%


3)

50% = 0.5

0.5 × ? = 50

divide each side of the equation by 0.5

? = 100


There you go! I really hope this helped, if there's anything just let me know! :)

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2 years ago
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The diameter of the circle below measures 7 cm. What is the approximate area of the circle?
GalinKa [24]
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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

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2 years ago
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sergejj [24]

Answer:

Well you can try to use a calculator, like the one on google. I can try to help if you really need it.

Step-by-step explanation:


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