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MAVERICK [17]
3 years ago
15

Sid wants to find the height of a tree without having to climb it, but it is a cloudy day so he cannot use the shadows . he take

s a mirror from his pocket and places it on thr ground 7.2m from the base of a tree. He backs up until he can see the top of the tree in the mirror, a distance of 1.2m from the mirror. If sids eyes are 1.5m above the ground what is the height of the tree?
Mathematics
2 answers:
devlian [24]3 years ago
8 0

Answer: Height of the tree is 9 m.

Step-by-step explanation:

Since we have given that

Distance at where he takes a mirror and placed it on the ground from the base of the tree = 7.2 m

Length of the tree from the mirror = 1.2 m

Distance at which he side eyes are above the ground = 1.5 m

Since there is direct variation between the actual height and the height in the mirror.

Let the height of the tree be h.

According to question,

\frac{1.5}{1.2}=\frac{x}{7.2}\\\\x=\frac{1.5\times 7.2}{1.2}\\\\x=9\ m

Hence, height of the tree is 9 m.

lozanna [386]3 years ago
5 0
The height of the tree would be 

1.5/1.2=x/7.2

x=9
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If Professor Wilson found that the test scores of his students had a variance of 4.4, what is the standard deviation? Type a num
malfutka [58]

Answer:

The standard deviation will be: 2.1        

Step-by-step explanation:

We know that standard deviation is basically the square root of variance.

Using the formula to calculate the standard deviation

s=\sqrt{s^2}

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so the standard deviation can be calculated as:

standard deviation =\sqrt{s^2}

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A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down t
VLD [36.1K]

Answer:

6000 ft

Step-by-step explanation:

Let length of rectangular field=x

Breadth of rectangular field=y

Area of rectangular field=1500000 square ft

Area of rectangular field=l\times b

Area of rectangular field=x\time y

1500000=xy

y=\frac{1500000}{x}

Fencing used ,P(x)=x+x+y+y+y=2x+3y

Substitute the value of y

P(x)=2x+3(\frac{1500000}{x})

P(x)=2x+\frac{4500000}{x}

Differentiate w.r.t x

P'(x)=2-\frac{4500000}{x^2}

Using formula:\frac{dx^n}{dx}=nx^{n-1}

P'(x)=0

2-\frac{4500000}{x^2}=0

\frac{4500000}{x^2}=2

x^2=\frac{4500000}{2}=2250000

x=\sqrt{2250000}=1500

It is always positive because length is always positive.

Again differentiate w.r.t x

P''(x)=\frac{9000000}{x^3}

Substitute x=1500

P''(1500)=\frac{9000000}{(1500)^3}>0

Hence, fencing is minimum at x=1 500

Substitute x=1 500

y=\frac{1500000}{1500}=1000

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Substitute the values

Shortest length of fence used=2(1500)+3(1000)=6000 ft

Hence, the shortest length of fence that the rancher can used=6000 ft

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4 years ago
7 qt. of a 27% saline solution was mixed
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Answer:

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So, the concentration of the mixture is 24% saline

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