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olga55 [171]
3 years ago
6

Given quadrilateral RSTU, determine if each pair of sides (if any) are parallel and which are perpendicular for the coordinates

of the vertices.
R(1, -3), S(4, -1), T(2, 2), U(-4, -2)

Click on the graphic to complete the answer.
Mathematics
1 answer:
Mrac [35]3 years ago
3 0
Slope RS: 2/3
Slope ST: -3/2
Slope TU: 4/6 = 2/3
Slope UR: -1/5

Sides RS and TU have the same slope, so they are parallel.
The slope of side ST is the negative reciprocal of the slopes of sides RS and TU, so side ST is perpendicular to sides RS and TU.
Side UR is not parallel or perpendicular to any other side.

The quadrilateral is a trapezoid.
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I already found the vertex, but can someone help me find the latus rectum of this parabola?
Fofino [41]

Answer:

Step-by-step explanation:

y=x²+7

vertex(0,7)

the length of latus rectum in a parabola equal to four times the focal length :

y=x²+7

focus X=-b/2a=0

focus Y=c- (b²-1)/4a=7+1/4=29/4

focus (0 , 29/4)

latus rectum is 29/4

(x-h)^2 = 4p (y-k)   4p is the length of the latus rectum with vertex(0,7)

(0-0)²=4p(29/4-7)

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100 Points and Brainly.
user100 [1]

function :  y = (-x) - 6

<u>Find x-intercept</u> :

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  • x = -6

<u>Find y-intercept</u> :

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mark these two points on both the axis and draw a straight linear graph.

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How would I do the steps to solve this?
allsm [11]

Answer:

The maximum revenue is 16000 dollars (at p = 40)

Step-by-step explanation:

One way to find the maximum value is derivatives. The first derivative is used to find where the slope of function will be zero.

Given function is:

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Taking derivative wrt p

\frac{d}{dp} (R(p) = \frac{d}{dp} (-10p^2+800p)\\R'(p) = -10 \frac{d}{dp} (p^2) +800 \ frac{d}{dp}(p)\\R'(p) = -10 (2p) +800(1)\\R'(p) = -20p+800\\

Now putting R'(p) = 0

-20p+800 = 0\\-20p = -800\\\frac{-20p}{-20} = \frac{-800}{-20}\\p = 40

As p is is positive and the second derivative is -20, the function will have maximum value at p = 40

Putting p=40 in function

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The maximum revenue is 16000 dollars (at p = 40)

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