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IrinaVladis [17]
3 years ago
9

According to the History 202 syllabus, the first paper is worth 25% of the course grade, the second paper is worth 20% of the co

urse grade, and the final exam is worth 55% of the overall course grade. Inara earns a 74% on the first term paper, an 99% on the second term paper, and the final exam has not happened yet. What is the lowest grade that Inara would need on the final exam in order to earn at least an 80% in the course?
Mathematics
1 answer:
OleMash [197]3 years ago
8 0

Answer:

  76%

Step-by-step explanation:

Let f represent the score on the final. Then Inara wants the final grade to be ...

  0.25·74 +0.20·99 +0.55·x ≥ 80

  38.3 +0.55x ≥ 80

  0.55x ≥ 41.7

  x ≥ 41.7/0.55 ≈ 75.82

Inara needs a final exam grade of 76% or better to earn at least 80% in the course.

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alina1380 [7]
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3x over 10 = 1,290 <br> please help
MAVERICK [17]

3x/10= 1290

Mutiply by 10 for both sides

3x/10*10= 1290*10

Cross out 10 and 10 , divide by 10 and then becomes 3x

3x=12900

divide by 3 for both sides

3x/3= 12900/3

Cross out 3 and 3 , divide by 3 and then becomes x

x= 4300

Answer: x= 4300

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4 years ago
The image of a point after a –90° rotation about the origin is (–4, 7). What are the coordinates of the preimage?.
ahrayia [7]
The coordinated of the preimage would be :

x  = x cos θ + y sin θ

y = x sin θ + y cos θ

where 

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4 years ago
MATH HELP PLZ!!!
RoseWind [281]

Answer:

a)    tan (157.5) = \frac{1-cos 315}{sin315}

b)

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c)

      sin^{2} (157.5) = \frac{1-cos (315) }{2}

d)

  cos 330° = 1- 2 sin² (165°)

       

         

Step-by-step explanation:

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

By using trigonometry formulas

a)

cos2∝  = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ =  2 cos² ∝/2

cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}

b)

cos2∝  = 1- 2 sin² ∝

cos∝  = 1- 2 sin² ∝/2

sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

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

Given

              tan\alpha = \frac{sin\alpha }{cos\alpha }

          we know that trigonometry formulas

        sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

         1- cos∝ =  2 sin² ∝/2

      Given

         tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }

put ∝ = 315

      tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }

     multiply with ' 2 sin (∝/2) both numerator and denominator

        tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2}  }{2sin(\frac{315}{2} cos(\frac{315}{2}) }

Apply formulas

 sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

  1- cos∝ =  2 sin² ∝/2

now we get

 tan (157.5) = \frac{1-cos 315}{sin315}

       

b)

          sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 330° above formula

             sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}

            sin^{2} (165) = \frac{1-cos (330) }{2}

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c )

         sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 315° above formula

             sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}

            sin^{2} (157.5) = \frac{1-cos (315) }{2}

           

d)

     cos∝  = 1- 2 sin² ∝/2

   put      ∝ = 330°

       cos 330 = 1 - 2sin^{2} (\frac{330}{2} )

      cos 330° = 1- 2 sin² (165°)

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3 years ago
Which problem solving strategy would be best to solve this problem?
bezimeni [28]

Answer:

b

Step-bystep explanation:

when subtracting you could reverse it and add so add back the 2.50 to the 5 and if he spent half of what he had left the that would mean he had another 5 dollars. in total he had 12.50$

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