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Kryger [21]
3 years ago
11

Which will result in a difference of squares?

Mathematics
1 answer:
Solnce55 [7]3 years ago
7 0

Answer:

Option C) is correct

That is the result should be in a difference of squares is

(-7x+4)(-7x-4)=(7x)^2-4^2

Step-by-step explanation:

To find the result in the form of differences in squares

Given that the expression is (-7x+4)(-7x-4)

(-7x+4)(-7x-4)=(-7x)^2-4^2 ( by using the identity (a+b)(a-b)=a^2-b^2 )

Here a=-7x and b=4

=(7x)^2-4^2

Therefore (-7x+4)(-7x-4)=(7x)^2-4^2 which results as a difference of squares.

Therefore Option C) is correct

That is the result should be in a difference of squares is

(-7x+4)(-7x-4)=(7x)^2-4^2

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TiliK225 [7]

Answer:

53

Step-by-step explanation:

complementary angle meaning it needs to have the whole sum of 90 degree.

90-37=53

8 0
4 years ago
Which of the following choices best justifies the conclusion based on the given information?
Sindrei [870]
To abide by the mathematical order of operations in this equation you must distribute "a" into "b" and "c" by multiplying, which is the distributive property. So the answer would be C. Distributive
8 0
3 years ago
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1,2,3,4,5,6,7,8,9,10 solve plz thank you
GalinKa [24]
1. 4x=64
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n<15
6. x-10=30
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8 0
4 years ago
22.Island A is 160 miles from island B. A ship captain travels 170 miles from island A and then finds that he is off course and
bixtya [17]

we know that

Applying the law of cosines: 


c² = a² + b² - 2abcos(C) 

where: 

a,b and c are sides of the triangle and C is the angle opposite side c 

let

a=170 mi

b=200 mi

c=160 mi

that is 

160² = 170² + 200² - 2(170)(200)cos(C) 


solve for C 


25,600 = 28,900 + 40,000 - 68,000cos(C) 

25,600 - 28,900 - 40,000 = -68,000cos(C) 

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C=arc cos(0.6367)--------> C=50.45°


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129.55 degrees

6 0
4 years ago
Prove the following
fomenos

Answer:

Step-by-step explanation:

\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

\rm \: {cos \: (2\pi + x) }

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx

\rm \: cot(2\pi + x) \: = \: cotx

So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}

\rm \: = \: 1=1

<h2>Hence,</h2>

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

  • sin (90°-θ)  =  cos θ
  • cos (90°-θ)  =  sin θ
  • tan (90°-θ)  =  cot θ
  • csc (90°-θ)  =  sec θ
  • sec (90°-θ)  =  csc θ
  • cot (90°-θ)  =  tan θ
  • sin (90°+θ)  =  cos θ
  • cos (90°+θ)  =  -sin θ
  • tan (90°+θ)  =  -cot θ
  • csc (90°+θ)  =  sec θ
  • sec (90°+θ)  =  -csc θ
  • cot (90°+θ)  =  -tan θ
  • sin (180°-θ)  =  sin θ
  • cos (180°-θ)  =  -cos θ
  • tan (180°-θ)  =  -tan θ
  • csc (180°-θ)  =  csc θ
  • sec (180°-θ)  =  -sec θ
  • cot (180°-θ)  =  -cot θ
  • sin (180°+θ)  =  -sin θ
  • cos (180°+θ)  =  -cos θ
  • tan (180°+θ)  =  tan θ
  • csc (180°+θ)  =  -csc θ
  • sec (180°+θ)  =  -sec θ
  • cot (180°+θ)  =  cot θ
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  • cos (270°-θ)  =  -sin θ
  • tan (270°-θ)  =  cot θ
  • csc (270°-θ)  =  -sec θ
  • sec (270°-θ)  =  -csc θ
  • cot (270°-θ)  =  tan θ
  • sin (270°+θ)  =  -cos θ
  • cos (270°+θ)  =  sin θ
  • tan (270°+θ)  =  -cot θ
  • csc (270°+θ)  =  -sec θ
  • sec (270°+θ)  =  cos θ
  • cot (270°+θ)  =  -tan θ
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3 years ago
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