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pickupchik [31]
3 years ago
11

rectangle has vertices at (Negative3, 2), (7, 2), (7, Negative 5), and (Negative 3, Negative 5). Jordan says the area of the rec

tangle is 70 square units. Steps Jordan’s Work Step 1 Base: StartAbsoluteValue negative 3 EndAbsoluteValue + StartAbsoluteValue 75 EndAbsoluteValue = 10 Step 2 Height: StartAbsoluteValue negative 5 EndAbsoluteValue + StartAbsoluteValue 2 EndAbsoluteValue = 7 Step 3 Area: 10 times 7 = 70 square units Which is true of Jordan’s solution? Jordan is not correct because he did not find the absolute value before multiplying the dimensions. Jordan is not correct because he did not add correctly before multiplying the dimensions. Jordan correctly found the area of the rectangle by adding the base and the height. Jordan correctly found the area of the rectangle by multiplying the base and the height.

Mathematics
2 answers:
Scilla [17]3 years ago
8 0

Answer:

Jordan correctly found the area of the rectangle by multiplying the base and the height.

Step-by-step explanation:

n200080 [17]3 years ago
6 0

Answer:

Jordan correctly found the area of the rectangle by multiplying the base and the height.

Step-by-step explanation:

Plot the vertices to better understand the problem

Let

A(-3,2),B(7,2),C(7,-5),D(-3,-5)

see the attached figure

we know that

The area of rectangle is equal to

A=bh

we have

b=AB\\h=BC

step 1

Find the base of rectangle

b=\left|7-\left(-3\right)\right|=10\ units

step 2

Find the height of rectangle

h=\left|2-\left(-5\right)\right|=7\ units

step 3

Find the area

A=bh

substitute the values

A=(10)(7)=70\ units^2

therefore

Jordan correctly found the area of the rectangle by multiplying the base and the height.

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hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to

Answer:  v = 5v + 4u + 1.5sin(3t),

  • 0
  • 1
  •  4
  • 5
  • 0
  • 1.5sin(3t)
  • 1
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Step-by-step explanation:

u" - 5u' - 4u = 1.5sin(3t)        where u'(1) = 2.5   u(1) = 1

v represents the "velocity function"   i.e   v = u'(t)

As v = u'(t)

<em>u' = v</em>

since <em>u' = v </em>

v' = u"

v'  = 5u' + 4u + 1.5sin(3t)   ( given that u" - 5u' - 4u = 1.5sin(3t) )

    = 5v + 4u + 1.5sin(3t)  ( noting that v = u' )

so v' = 5v + 4u + 1.5sin(3t)

d/dt \left[\begin{array}{ccc}u&\\v&\\\end{array}\right]= \left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]  \left[\begin{array}{ccc}u&\\v&\\\end{array}\right] + \left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]

Given that u(1) = 1 and u'(1) = 2.5

since v = u'

v(1) = 2.5

note: the initial value for the vector valued function is given as

\left[\begin{array}{ccc}u(1)&\\v(1)\\\end{array}\right]  = \left[\begin{array}{ccc}1\\2.5\\\end{array}\right]

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