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1 year ago

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The distance d (in miles) that you can see to the horizon with your eye level h feet above the water is given by d=3h2−−−√. You

are standing on a platform. Your eye level is 12 feet above the water. Can you see a ship that is 4.5 miles away? because you can see about miles.

1 answer:

liubo4ka [24]1 year ago

7 0

Answer: He cannot see a ship that is 4.5 miles away.

Step-by-step explanation:

Since we have given that

$d=3\sqrt{h}$

Here, h is the eye level above the water

d is the distance away.

If h = 12 feet

We will get :

$d=3\sqrt{12}\\\\d=3\times 2\sqrt{3}\\\\d=6\sqrt{3}\\\\d=10.39$

Hence, it is 10.39 feet away when eye level is 12 feet above the water.

Therefore , he cannot see a ship that is 4.5 miles away.

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Answer:

(a) $Perimeter = 32.2\ cm$

(b) 100

Step-by-step explanation:

Solving (a):

Given

Shape: Rectangle

$Area = 64.8$

Required

Calculate the perimeter.

Area is calculated as:

$Area = L * W$

Where

$L = Length$ and $W = Width$

Substitute 64.8 for Area

$64.8 = L * W$

Make L the subject:

$L = \frac{64.8}{W}$

Perimeter is calculated as:

$P = 2 * (L + W)$

Substitute 64.8/W for L

$P = 2 * (\frac{64.8}{W} + W)$

$P = \frac{129.6}{W} + 2W$

To solve further, we take the derivative of P and set it to 0, afterwards.

$dP = -\frac{129.6}{W^2} + 2$

Set to 0

$0 = -\frac{129.6}{W^2} + 2$

Collect Like Terms

$\frac{129.6}{W^2} = 2$

Cross Multiply:

$2W^2= 129.6$

Divide through by 2

$W^2 = 64.8$

Take square roots

$W = \sqrt{64.8$

$W = 8.05$

Recall that:

$L = \frac{64.8}{W}$

So:

$L= \frac{64.8}{8.05}\\$

$L= 8.05$

The perimeter is:

$Perimeter = 2 * (8.05 + 8.05)$

$Perimeter = 2 * (16.10)$

$Perimeter = 32.2\ cm$

Solving (b):

Given

((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)

Required

Solve

$Tenths = \frac{1}{10}$

So, the expression becomes:

((1 Group of $10 * \frac{1}{10}$ ) and (1 group of $8 * \frac{1}{10}$ ))/(6 groups of $3 * \frac{1}{10}$ )

This gives:

((1 Group of $\frac{10}{10}$ ) and (1 group of $\frac{8}{10}$ ))/(6 groups of $\frac{3}{10}$ )

Group means product, so the expression becomes:

$\frac{(1 * \frac{10}{10} \ and\ 1 * \frac{8}{10})}{6 * \frac{3}{10}}$

And, as used here means addition

$\frac{(1 * \frac{10}{10} + 1 * \frac{8}{10})}{6 * \frac{3}{10}}$

Simplify:

$\frac{(1 * 1 + 1 * 0.80)}{6 *0.30}$

$\frac{(1 + 0.80)}{1.80}$

$\frac{1.80}{1.80}$

$= 1.00$

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