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Sliva [168]
3 years ago
13

I NEED URGENT HELP WITH 18-21

Mathematics
2 answers:
Genrish500 [490]3 years ago
4 0

Answer:

. . . 18. x = 8√6

. . . 19. x = 3

. . . 20. x = 3√3

. . . 21. x = 5/3

Step-by-step explanation:

You have already marked the correct relationships on the drawings. Finding the value of x is a matter of making use of those relationships.

__

18. As you show, x is the given side multiplied by √3, so is ...

. . .x = (8√2)(√3) = 8√6

__

19. As you show, the given side is x multiplied by √2, so ...

. . . 3√2 = x√2

. . . 3 = x . . . . . . . . . . divide by √2

__

20. As you show, the given side is x multiplied by √3, so ...

. . . x√3 = 9

. . . x = 3√3 . . . . . . multiply by (√3)/3

__

21. As you show, ...

. . . 8x -10 = 2x

. . . 6x -10 = 0 . . . . . subtract 2x

. . . x - 10/6 = 0 . . . . divide by 6

. . . x = 5/3 . . . . . . . . reduce the fraction and add it to both sdes

Sholpan [36]3 years ago
3 0
18) x = 8√6
19) x = (3√2)/2
20) x = 3√3
21) roots of x are x= 1 or x = 5/3
x can't be 1 so x = 5/3

Tip for you: You are not meant to use Sine or Cosine Laws/Rules

What I used to obtain my answers
• Pythagorean Theorem
• Trigonometric ratios
• Quadratic equation
• Linear algebra
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Answer:

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Step-by-step explanation:

According to the Question,

Given, A circle with centre F, ∠EFG=54 and EF=19 .

length of arc EG = Radius(EF) × ∠EFG(in Radian)

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(For Diagram please find in attachment)

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\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: f(0) = -2

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \: 2 {x}^{2}  - 3y = 6

\qquad \tt \rightarrow \: 3y + 6 = 2 {x}^{2}

\qquad \tt \rightarrow \: 3y = 2 {x}^{2}  - 6

\qquad \tt \rightarrow \: y =  \cfrac{2 {x}^{2} - 6 }{3}

\qquad \tt \rightarrow \: y =  \cfrac{2 {}^{} }{3}  {x}^{2}  - 2

[ here, y can be replaced with f(x) because y is a function of x ]

\qquad \tt \rightarrow \: f(x) =  \cfrac{2 {}^{} }{3}  {x}^{2}  - 2

\large\textsf{Find f(0):}

\qquad \tt \rightarrow \: f(0) =  \cfrac{2 {}^{} }{3}  {(0)}^{2}  - 2

\qquad \tt \rightarrow \: f(0) =  0  - 2

\qquad \tt \rightarrow \: f(0) =   - 2

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