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just olya [345]
4 years ago
15

For what values of x and y is PQRS and parallelogram

Mathematics
1 answer:
LiRa [457]4 years ago
5 0

Answer:

x = 2

y = 3

Step-by-step explanation:

Measures of opposite sides of a parallelogram are equal.

Therefore,

PQ = SR

2y + 2 = y + 5

2y - y = 5 - 2

y = 3

QR = PS

2x + 3 = y + 4

2x + 3 = 3 + 4 (y = 3)

2x = 7 - 3

2x = 4

x = 4/2

x = 2

Thus, for x = 2 and y = 3, PQRS is a parallelogram.

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Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisti
timofeeve [1]

Answer:

0

Step-by-step explanation:

We want to compute the curve integral (or line integral)

\bf \int_{C}F

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write  

C = \bf C_1+C_2

where  

\bf C_1 =  line segment from (1, 5) to (0, 0)  

\bf C_2 = line segment from (0, 0) to (0, 9)

so,

\bf \int_{C}F=\int_{C_1}F+\int_{C_2}F

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

<em>r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1 </em>

Hence \bf C_1 can be parameterized as

\bf r_1(t) = (1-t, 5-5t) for 0 ≤ t ≤ 1

and \bf C_2 can be parameterized as

\bf r_2(t) = (0, 9t) for 0 ≤ t ≤ 1

The derivatives are

\bf r_1'(t) = (-1, -5)

\bf r_2'(t) = (0, 9)

and

\bf \int_{C_1}F=\int_{0}^{1}F(r_1(t))\circ r_1'(t)dt=\int_{0}^{1}(5-5t,t-1)\circ (-1,-5)dt=0

\bf \int_{C_2}F=\int_{0}^{1}F(r_2(t))\circ r_2'(t)dt=\int_{0}^{1}(9t,0)\circ (0,-9)dt=0

In consequence,

\bf \int_{C}F=0

6 0
4 years ago
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Airida [17]

Answer:

x = 17

Step-by-step explanation:

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2x - 3x = -7 - 10 (collecting like terms)

- x = -17

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x = 17

4 0
3 years ago
87-(36.5-11.5)*(-3)+9 simplify using order of operations
OLga [1]
Remember PEMDAS 
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4 years ago
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Mars2501 [29]

Answer:

x=-4

Step-by-step explanation:

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