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3 years ago

Ex) A nursery plants a new tree and attaches a guy wire to help support the tree

Mathematics

2 answers:

o-na [289]3 years ago

8 0

Answer:

d = 5,35 ft

Step-by-step explanation: See annex

Figure in annex is clear,

sin ∠42⁰ = d / 8

And

sin ∠42⁰ = 0.669

d/8 = 0,669

d = 0,669*8

d = 5,35 ft

nordsb [41]3 years ago

6 0

Answer:

a ≈ 5.4 ft

Step-by-step explanation:

A guy wire is used to hold a tree firmly to the ground to avoid bending or falling due to earth natural phenomenon like wind.

The guy wire is tied to a stake and attached to the tree. The illustration above forms a right angle triangle. The angle of elevation from the stake in the ground with the tree is 42°.

The tree forms the opposite side of the right angle triangle starting from the point the guy wire was attached down to the base of the tree. The guy wire attached from the tree to the stake forms the hypotenuse of the right angle triangle.

The height of the connection point on the tree is the height of the tree starting from where the guy wire was tied to the tree to the base. Let us call the height a.

hypotenuse = 8 ft

opposite side(a) = ?

angle of elevation = 42°

Using the SOHCAHTOA principle

sin ∅ = opposite/hypotenuse

sin 42° = a/8

cross multiply

a = 8 sin 42°

a = 8 × 0.6691306064

a = 5.3530448509

a ≈ 5.4 ft

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Anestetic [448]

Answer:

$x=\dfrac{-17+ \sqrt{451} }{18}, \quad x=\dfrac{-17- \sqrt{451} }{18}$

Step-by-step explanation:

$\textsf{Given equation}:$

$\dfrac{2}{3}x+\dfrac{3x}{x}=\dfrac{4x}{2}(3x+6)$

$\textsf{Cancel the common factor } x \textsf{ in }\dfrac{3x}{x}:$

$\implies \dfrac{2}{3}x+3=\dfrac{4x}{2}(3x+6)$

$\textsf{Simplify }\dfrac{4x}{2}\textsf{ by dividing the numbers}:$

$\implies \dfrac{2}{3}x+3=2x(3x+6)$

$\textsf{Apply the distributive law}\quad \:a\left(b+c\right)=ab+ac:$

$\implies \dfrac{2}{3}x+3=6x^2+12x$

$\textsf{Multiply both sides by 3 to cancel the denominator of }\dfrac{2}{3}:$

$\implies \dfrac{2}{3}x\cdot 3+3\cdot 3=6x^2\cdot 3+12x \cdot 3$

$\implies 2x+9=18x^2+36x$

$\textsf{Switch sides}:$

$\implies 18x^2+36x=2x+9$

$\textsf{Subtract }2x \textsf{ from both sides}:$

$\implies 18x^2+36x-2x=2x+9-2x$

$\implies 18x^2+34x=9$

$\textsf{Subtract 9 from both sides}:$

$\implies 18x^2+34x-9=9-9$

$\implies 18x^2+34x-9=0$

Solve using the Quadratic Formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Define the variables:

$\implies a=18, \quad b=34, \quad c=-9$

Substitute the defined variables into the quadratic formula and solve for x:

$\implies x=\dfrac{-34 \pm \sqrt{34^2-4(18)(-9)} }{2(18)}$

$\implies x=\dfrac{-34 \pm \sqrt{1156+648} }{36}$

$\implies x=\dfrac{-34 \pm \sqrt{1804} }{36}$

$\implies x=\dfrac{-34 \pm \sqrt{4 \cdot 451} }{36}$

$\implies x=\dfrac{-34 \pm \sqrt{4}\sqrt{451} }{36}$

$\implies x=\dfrac{-34 \pm 2\sqrt{451} }{36}$

$\implies x=\dfrac{-17 \pm \sqrt{451} }{18}$

Therefore, the solutions to the given equation are:

$x=\dfrac{-17+ \sqrt{451} }{18}, \quad x=\dfrac{-17- \sqrt{451} }{18}$

Learn more about the Quadratic Formula here:

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