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Butoxors [25]
3 years ago
9

Find x, the unknown side. A) 4.6 C) 4.8 B) 5.4 D)3.5

Mathematics
2 answers:
never [62]3 years ago
5 0

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Trigonometry.

==> here lets take, cos 33 = Adjacent / Hypotenuse

==> Cos 33° = x/5.5

hence, x = (5.5) * Cos 33°

==> x = 0.83 * 5.5

==> x = 4.61 units

hence, x = 4.6 units

MrMuchimi3 years ago
3 0
Go ahead to do the trick
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<u><em>Answer:</em></u>

108 square units

<h3><u><em>Step-by-step explanation:</em></u></h3>

3 x 4 = 12

3 x 4 = 12

6 x 3 = 18

6 x 3 = 18

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2 years ago
A The length of a rectangle is 4 m more
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Given :

  • The length of a rectangle is 4m more than the width.
  • The area of the rectangle is 45m²

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To Find :

  • The length and width of the rectangle.

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Solution :

We know that,

\qquad { \pmb{ \bf{Length \times Width = Area_{(rectangle)}}}}\:

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Let's assume the length of the rectangle as x and the width will be (x – 4).

⠀

Now, Substituting the given values in the formula :

\qquad \sf \: { \dashrightarrow x  \times  (x - 4) = 45 }

\qquad \sf \: { \dashrightarrow {x}^{2}  - 4x = 45 }

\qquad \sf \: { \dashrightarrow {x}^{2}  - 4x - 45 = 0 }

\qquad \sf \: { \dashrightarrow {x}^{2}    - 9x+ 5 x - 45 = 0 }

\qquad \sf \: { \dashrightarrow x(x - 9) + 5(x - 9) = 0 }

\qquad \sf \: { \dashrightarrow (x  - 9) (x  + 5) = 0 }

\qquad \sf \: { \dashrightarrow x = 9, \: \: x =  - 5}

⠀

Since, The length can't be negative, so the length will be 9 which is positive.

⠀

\qquad { \pmb{ \bf{ Length _{(rectangle)} = 9\:m}}}\:

\qquad { \pmb{ \bf{ Width _{(rectangle)} = 9 - 4=5\: m}}}\:

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