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Alex73 [517]
3 years ago
15

Function 1: y = 4x + 5

Mathematics
1 answer:
evablogger [386]3 years ago
6 0
Answer: function 1

Rate of change of function 1:
Following the format of y=mx+c, the rate of change should be m, so, the rate of change for function 1 = 4

To find the gradient (rate of change):
The two points the line passes through are (x1, y1) and (x2, y2), which in this case is (1, 6) and (3, 10)
(Doesn't matter which is which but you need to make sure that once you decide which is which, you stick to it)
To calculate the gradient, you substitute these values following (y1 - y2)/(x1 - x2)

Gradient of function 2 = (10 - 6)/(3 - 1)
= 2

Therefore, since 4 > 2, rate of change of function 1 > rate of change of function 2.
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A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of
Gwar [14]

\bold{\huge{\underline{ Solution}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • A marker in the center of the fairway is 150 yards away from the centre of the green
  • While standing on the marker and facing the green, the golfer turns 100° towards his ball
  • Then he peces off 30 yards to his ball

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>distance </u><u>between </u><u>the </u><u>golf </u><u>ball </u><u>and </u><u>the </u><u>center </u><u>of </u><u>the </u><u>green </u><u>.</u>

<h3><u>Let's </u><u> </u><u>Begin </u><u>:</u><u>-</u></h3>

Let assume that the distance between the golf ball and central of green is x

<u>Here</u><u>, </u>

  • Distance between marker and centre of green is 150 yards
  • <u>That </u><u>is</u><u>, </u>Height = 150 yards
  • For facing the green , The golfer turns 100° towards his ball
  • <u>That </u><u>is</u><u>, </u>Angle = 100°
  • The golfer peces off 30 yards to his ball
  • <u>That </u><u>is</u><u>, </u>Base = 30 yards

<u>According </u><u>to </u><u>the </u><u>law </u><u>of </u><u>cosine </u><u>:</u><u>-</u>

\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}

  • Here, a = perpendicular height
  • b = base
  • c = hypotenuse
  • cos theta = Angle of cosine

<u>So</u><u>, </u><u> </u><u>For </u><u>Hypotenuse </u><u>law </u><u>of </u><u>cosine </u><u>will </u><u>be </u><u>:</u><u>-</u>

\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}

\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}

\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}

\sf{ x^{2} = 22500 + 900 + 156.6}

\sf{ x^{2} = 23556.6}

\bold{ x = 153.48\: yards }

Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards

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Answer:

Step-by-step explanation:

So here we have a 45-45-90 triangle.

This a special right triangle were the sides across from the 45 degree angles can be considered x, while the hypotenuse is two square roots of x.

Here since we have the sides across from the 45 degree angle we can conclude that x = 9\sqrt{2}

So if we wanted the hypotenuse we would just plug in this value of x like so:

hyp = x\sqrt{2} \\

hyp = (9\sqrt{2})(\sqrt{2})

hyp = 9(2)\\

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Therefore the hypotenuse is 18.

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