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fiasKO [112]
3 years ago
13

Lucy's final exam has true/false questions, worth 2 points each, and multiple choice questions, worth 3 points each. Let x be th

e number of true/false questions she gets correct, and let y be the number of multiple choice questions she gets correct. She needs more than 90 points on the exam to get an A in the class. Using the values and variables given, write an inequality describing this.
Mathematics
2 answers:
svlad2 [7]3 years ago
5 0
2x + 3y > 90

Use the term LEG.

L - Least
E-equal
G- Greater
Bad White [126]3 years ago
4 0
True/False questions = 2 points each(x).
Multiple choice questions = 3 points each(y).
She needs more than (>) 90 points.

2x + 3y > 90
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Determine the open t-intervals on which the curve is concave downward or concave upward. (Enter your answer using interval notat
Rashid [163]

Solution :

We have been given a parametric curve :

x = sin t , y = cos t , 0 < t < π

In order to determine concavity of the given parametric curve, we need to evaluate its second derivative first.

Therefore,

$\frac{dx}{dt} = \cos t, \ \frac{dy}{dt}= - \sin t$

$\therefore \frac{dy}{dx}= \frac{- \sin t}{\cos t}$

        $=- \tan t$

Taking double derivatives of the above equation:

$\frac{d^2y}{dx^2}= - \frac{d}{dx}(\tan t) $

      $= - \sec^2 t \frac{dt}{dx}$

     $= - \sec^2 t \left(\frac{1}{\cos t}\right)$

    $= - \sec^3 t$

For the concave up, we have

$\frac{d^2y}{dx^2} > 0$

$\Rightarrow - \sec^3 t > 0$

∴ $t \  \epsilon \left( \frac{\pi }{2}, \pi \right)$

For the concave down, we have

$\frac{d^2y}{dx^2} < 0$

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8 0
3 years ago
Solve the system by using elementary row operations on the equations. Follow the systematic elimination procedure
Umnica [9.8K]

Answer:

The solution of the Given matrix

  ( x₁ ,    x ₂ ) = ( - 5 , 4 )

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given equations are  x₁+4 x₂ = 11 ...(i)

                                 2 x₁ + 7 x₂= 18 ...(ii)

The matrix form

                                A X = B

            \left[\begin{array}{ccc}1&4\\2&7\\\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}11\\8\\\end{array}\right]

<u>Step(ii):-</u>

      \left[\begin{array}{ccc}1&4\\2&7\\\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}11\\8\\\end{array}\right]

The Augmented Matrix form is

[AB] = \left[\begin{array}{ccc}1&4&11\\2&7&18\\\end{array}\right]

Apply Row operations,  R₂ → R₂-2 R₁

[AB] = \left[\begin{array}{ccc}1&4&11\\0&-1&-4\\\end{array}\right]  

The matrix form

                   \left[\begin{array}{ccc}1&4\\0&-1\\\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}11\\-4\\\end{array}\right]

The equations are

                     x₁ + 4 x₂ = 11 ...(a)

                         - x ₂ = - 4

                            x ₂ = 4

Substitute   x ₂ = 4 in equation (a)

                      x₁ + 4 x₂ = 11

                      x₁ = 11 - 16

                        x₁ = -5

<u>Final answer</u>:-

The solution of the Given matrix

  ( x₁ ,    x ₂ ) = ( - 5 , 4 )

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