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adell [148]
3 years ago
5

The law of cosines is a2 +62 - 2abcosC = (2. Find the value of 2abcosC.

Mathematics
2 answers:
iren2701 [21]3 years ago
4 0

Answer:

A. 37

Step-by-step explanation:

azamat3 years ago
3 0

Answer:

2abcosC = 37

Step-by-step explanation:

Given

a^2 + b^2 - 2abcosC = c^2

Required

2abcosC

a^2 + b^2 - 2abcosC = c^2

Add 2abcosC to both sides

a^2 + b^2 - 2abcosC+2abcosC = c^2+2abcosC

a^2 + b^2 = c^2+2abcosC

Subtract c^2 from both sides

a^2 + b^2 -c^2= c^2 -c^2+2abcosC

a^2 + b^2 -c^2= 2abcosC

From the attachment:

a = 4 b = 5 and c = 2

So, we have:

4^2 + 5^2 -2^2= 2abcosC

16 + 25 -4= 2abcosC

37= 2abcosC

i.e.

2abcosC = 37

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-2(7a+ 9b) = –14a + [?]b
OverLord2011 [107]

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-18

Step-by-step explanation:

Left Hand Side = -2(7a + 9b) = -14a - 18b

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3 0
3 years ago
Read 2 more answers
What is the value of the expression below when w = 3. 4w^2 - 7w - 8
Makovka662 [10]
<h2>Answer:<em> </em><em><u>w =(-40-√4320)/-34=(20+6√ 30 )/17= 3.110 </u></em></h2><h2><em><u> w =(-40+√4320)/-34=(20-6√ 30 )/17= -0.757</u></em></h2>

Step-by-step explanation:  The prime factorization of  4320   is

  2•2•2•2•2•3•3•3•5  

To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 4320   =  √ 2•2•2•2•2•3•3•3•5   =2•2•3•√ 30   =

               ±  12 • √ 30

 √ 30   , rounded to 4 decimal digits, is   5.4772

So now we are looking at:

          w  =  ( -40 ± 12 •  5.477 ) / -34

Two real solutions:

w =(-40+√4320)/-34=(20-6√ 30 )/17= -0.757

or:

w =(-40-√4320)/-34=(20+6√ 30 )/17= 3.110

MY HEAD HURTS!

3 0
3 years ago
Answer using set notation if possible.
kirill115 [55]

Answer:

Domain:

[-5,-2) U (-2,3.5) U (3.5,5]

Range:

[-3 0) U (0,5]

It is not a function because x = 1 has two different outputs (3 & 5)

8 0
3 years ago
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