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hichkok12 [17]
2 years ago
6

Please help meeeeeeeeeeeeeeeeeeeeeeeeeeeee

Mathematics
1 answer:
natulia [17]2 years ago
8 0

Answer:

77 m (nearest metre)

Step-by-step explanation:

<u>Angle of Elevation</u>

If a person stands at a point and looks up at an object, the angle between their <u>horizontal line of sight</u> and the <u>object</u> is called the angle of elevation.

To find how tall the flagpole is, model as a right triangle and solve using the tan trigonometric ratio.

<u>Tan trigonometric ratio</u>

  \sf \\tan(\theta)=\dfrac{O}{A}

where:

  • \theta is the angle
  • O is the side opposite the angle
  • A is the side adjacent the angle

Given information:

  • Person height = 1.6 m
  • Horizontal distance (from person to flagpole) = 35 m
  • Angle of elevation = 65°

Therefore:

  • \theta = 65°
  • O = height of flagpole <u>less</u> height of person
  • A = 35 m

As the person is 1.6 m tall, we model the line of sight as 1.6 m <u>above ground level</u>.  Therefore, when calculating the height of the flagpole, we will need to add the height of the person to the value found using the trig ratio.

Substitute the given values into the formula and solve for O:

\implies \sf \tan(65^{\circ})=\dfrac{O}{35}

\implies \sf O=35\tan(65^{\circ})

Therefore the height of the flagpole is:

\implies \sf 35\tan(65^{\circ})+1.6=76.65774222...

\implies \sf Height\:of\:flagpole=77\:m\:(nearest\:metre)

Learn more about Angles of Elevation here:

brainly.com/question/27933160

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Answer:

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Step-by-step explanation:

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10x + 15 + 5x = 180, that is

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15x = 165 ( divide both sides by 15 )

x = 11

Thus

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Step-by-step explanation:




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Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x
fredd [130]

Answer:

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

6\cdot x^{2} \geq 10 + 11\cdot x

6\cdot x^{2}-11\cdot x -10 \geq 0

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Roots are found by Quadratic Formula:

r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}

r_{1} = \frac{5}{2} and r_{2} = -\frac{2}{3}

Then, the factorized form of the inequation is:

6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0

By Real Algebra, there are two condition that fulfill the inequation:

a) x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0

x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}

x \geq \frac{5}{2}

b) x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0

x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}

x\leq -\frac{2}{3}

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

3 0
3 years ago
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