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Eva8 [605]
4 years ago
8

Write the equation of the line parallel to y = 2x + 1 that passes through the point (6, 2).

Mathematics
1 answer:
stellarik [79]4 years ago
6 0

Answer:

y=2x-10

Step-by-step explanation:

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Can someone help me with this?!
NNADVOKAT [17]

Answer:

\sqrt{0/9}

Step-by-step explanation:

Have a good day :)

4 0
3 years ago
ASAP someone please use the Pythagorean Theorem to solve for the missing side of these two right triangles.
solong [7]

Answer:

9) x = 5

10) x = 4

Step-by-step explanation:

a^2 + b^2 = c^2

9) 6^2 + x^2 = 8^2

  - 36 + x^2 = 64

  -  36            - 36

_______________

           x^2 = 28

Square root them

Squareroot x^2 is x

Square root 28 is 5.29 round is 5

so x = 5

*Correct me if im wrong*

10) 3^2 + x^2 = 5^2

     9 + x^2 = 25

  -  9              - 9

____________________

             x^2= 16

Square root them

square root of x ^2 is x and square root of 16 is 4

so x = 4

6 0
3 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
  • sin (270°-θ)  =  -cos θ
  • 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
7 0
3 years ago
Read 2 more answers
For what value of x is the value of f(x)=4x-2 equal to 18
mina [271]
F(x) = 4(5) - 2 → f(x) = 20 - 2 → f(x) = 18
5 0
3 years ago
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I don't get the problem
Ulleksa [173]
A(b + c) = a*(b + c) = a*b + a*c

You must multiply individual terms and see what it would equal
3 0
3 years ago
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