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3 years ago

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Mathematics

2 answers:

andreev551 [17]3 years ago

8 0

23 miles

15-2.35 = 12.65
12.65/ .55= 23

Fittoniya [83]3 years ago

4 0

It has to be 23 miles fam

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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>Given :-</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

We have to find the distance between the golf ball and the center of the green .

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

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

Here, 

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

According to the law of cosine :-

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

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

So, For Hypotenuse law of cosine will be :-

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

Subsitute the required values,

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

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

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

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

$\sf{ x^{2} = 23556.6}$

$\bold{ x = 153.48\: yards }$

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

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2 years ago

A half-acre building lot is five times as long as it is wide. What are its dimensions?

Gnoma [55]

we know that

A half-acre is equal to----------> $2,023.43\ m^{2}$

Let

x---------> the length side of the building lot

y--------> the width side of the building lot

A------> Area of the building lot

x=5y-------> equation 1

$A=2,023.43\ m^{2}$

$A=x*y$

$2,023.43=x*y$ --------> equation 2

substitute equation 1 in equation 2

$2,023.43=[5y]*y$

$5y^{2}= 2,023.43\\y^{2}= 404.686\\y= 20.11\ m$

Find the value of x

$x=5y---> x=5*20.11--> x=100.55 m$

therefore

the answer is

the dimensions of the building lot are 100.55 m* 20.11 m

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Svetradugi [14.3K]

Answer:

i think 2-3 min.

Step-by-step explanation: i wish it's help u

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Please help!!!!!! + lots of points

Ilya [14]

Hey!

Hope this helps...

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The correct answers are A, B, D, and E

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