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dmitriy555 [2]
2 years ago
10

Please Help - Point x^3-4x<-5x^2+20 Solve in interval notation and list the zeros

Mathematics
1 answer:
Naddika [18.5K]2 years ago
4 0

hi my name is antonette Jane S pascua please help me my question thank you for your answer

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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

5 0
2 years ago
Find the length BC
Nady [450]
Answer:

The length of BC = 12

Step-by-step explanation:

c^2 = a^2 + b^2

15^2 = 9^2 + b^2

225 = 81 + b^2

b^2 = 225 - 81

b^2 = 144

b = 12

The length of BC is 12
5 0
3 years ago
0.75 recurring as a fraction<br> with 7 and 5 recurring
mariarad [96]

   100x = 75.757575...
  -      x =   0.757575...
 -----------------------------
     99x = 75

x = 75/99

x = 25/33

Answer: 0.757575... = 25/33
6 0
3 years ago
Which fraction (in simplest form) is equivalent to 0.03?<br> A.3/100<br> B.3/10<br> C.1/3<br> D.1/30
bija089 [108]
It is (A). When removing decimal of 0.03 you will get = 3/100 so A is the correct answer.
3 0
3 years ago
3. A researcher randomly selects a sample of 61 former student leaders from a list of graduates of UNCG who had participated in
Sophie [7]

Answer:

not statistically significant at ∝ = 0.05

Step-by-step explanation:

Sample size( n )  = 61

Average for student leader graduates to finish degree ( x') = 4.97 years

std = 1.23

Average for student body = 4.56 years

<u>Determine if the difference between the student leaders and the entire student population is statistically significant at alpha</u>

H0( null hypothesis ) : u = 4.56

Ha : u ≠ 4.56

using test statistic

test statistic ; t = ( x' - u ) / std√ n

                        = ( 4.97 - 4.56 ) / 1.23 √ 61

                        = 2.60

let ∝ = 0.05 , critical value = -2.60 + 2.60

Hence we wont fail to accept  H0

This shows that the difference between the student leaders and the entire student population is not statistically significant at ∝ = 0.05

3 0
3 years ago
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