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2 years ago

Simplify:$$\sqrt{2\sqrt{8^2+15^2}+\sqrt{9^2+40^2}}$$

Mathematics

2 answers:

bagirrra123 [75]2 years ago

8 0

Answer:

$5\sqrt{3}$

Step-by-step explanation:

$\sqrt{2\sqrt{8^2+15^2}+\sqrt{9^2+40^2}}=?\\\\1)\sqrt{8^2+15^2}=\sqrt{289}=17\\2)\sqrt{9^2+40^2}=\sqrt{1681}=41\\3)2\times17=34\\4)\sqrt{34+41}=5\sqrt{3}$

All Done!

Sati [7]2 years ago

4 0

Answer:

Your correct answer is 8.660254

Step-by-step explanation:

√2√82 + 152 + √92 + 402 = 8.660254

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The stopping distance d of an automobile is directly proportional to the square of its speed s. On one road, a car requires 75 f

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Answer:

The car requires 192 feet to stop from a speed of 48 miles per hour on the same road

Step-by-step explanation:

Direct proportion means that two quantities increase or decrease in the same ratio

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∵ The stopping distance d of an automobile is directly

proportional to the square of its speed s

- That means d ∝ s²

∴ $\frac{d_{1}}{d_{2}}=\frac{(s_{1})^{2}}{(s_{2})^{2}}$

∵ A car requires 75 feet to stop from a speed of 30 miles per hour

∴ d = 75 feet

∴ s = 30 miles/hour

- Change the mile to feet

∵ 1 mile = 5280 feet

∴ 30 miles/hour = 30 × 5280 = 158400 feet/hour

∵ The car require to stop from a speed of 48 miles per hour

on the same road

- Change the mile to feet

∴ 48 miles/hour = 48 × 5280 = 253440 feet/hour

∵ $\frac{d_{1}}{d_{2}}=\frac{(s_{1})^{2}}{(s_{2})^{2}}$

- Substitute the values of $d_{1}$ by 75 feet, $s_{1}$ by 158400 feet/hour

and $s_{2}$ by 253440 feet/hour

∴ $\frac{75}{d_{2}}=\frac{(158400)^{2}}{(253440)^{2}}$

∴ $\frac{75}{d_{2}}=\frac{25}{64}$

- By using cross multiplication

∴ 25 × $d_{2}$ = 75 × 64

- Divide both sides by 25

∴ $d_{2}$ = 192 feet

The car requires 192 feet to stop from a speed of 48 miles per hour on the same road

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