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2 years ago

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A rectangular parking lot is surrounded on its perimeter by a fence that is 1200 feet long. if the length of a rectangular parki

ng lot is 5 times it’s width which equation can be used to find the width of the parking lot

2 answers:

r-ruslan [8.4K]2 years ago

8 0

Answer:

fah)',55%9+*_)64•%π\¥)--_f

AfilCa [17]2 years ago

4 0

Answer:

2w+2(w+5)=1200

Step-by-step explanation:

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It takes Dan 32 minutes to complete 2 pages of math homework. At this rate, how many pages does he complete in 200 minutes?

Anna [14]

He can complete about 6 pages of math homework.
you first do 200 divided by 32 and you get 6.25. So therefore Dan can complete 6 pages in 200 minutes.

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X = [-(-8) +/- sqrt((-8)^2 -4*41)] / 2

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3 years ago

Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence. The angle

dem82 [27]

Answer:

229.23 feet.

Step-by-step explanation:

The pictorial representation of the problem is attached herewith.

Our goal is to determine the height, h of the tree in the right triangle given.

In Triangle BOH

$Tan 60^0=\dfrac{h}{x}\\h=xTan 60^0$

Similarly, In Triangle BOL

$Tan 50^0=\dfrac{h}{x+60}\\h=(x+60)Tan 50^0$

Equating the Value of h

$xTan 60^0=(x+60)Tan 50^0\\xTan 60^0=xTan 50^0+60Tan 50^0\\xTan 60^0-xTan 50^0=60Tan 50^0\\x(Tan 60^0-Tan 50^0)=60Tan 50^0\\x=\dfrac{60Tan 50^0}{Tan 60^0-Tan 50^0} ft$

Since we have found the value of x, we can now determine the height, h of the tree.

$h=\left(\dfrac{60Tan 50^0}{Tan 60^0-Tan 50^0}\right)\cdotTan 60^0\\h=229.23 feet$

The height of the tree is 229.23 feet.

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3 years ago

Answer the question in the picture

nekit [7.7K]

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$

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