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1 year ago

Find the length of the third side. If necessary, round to the nearest tenth.

Mathematics

2 answers:

viva [34]1 year ago

5 0

Answer:

the question states if necessary, round the nearest tenth and by using the pythagoras theorem the answer should be 10 since its 9.21 and its nearest 10th is 10? im not quite sure though

lina2011 [118]1 year ago

4 0

Answer:

≈ 9.2

Step-by-step explanation:

using Pythagoras' identity in the right triangle

the square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

let x be the third side ( hypotenuse ) , then

x² = 7² + 6² = 49 + 36 = 85 ( take square root of both sides )

x = $\sqrt{85}$ ≈ 9.2 ( to the nearest tenth )

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x = 112

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Please solve this (Rational numbers 8th grade)

pochemuha

Question # 1

Answer:

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}$

Step-by-step explanation:

Given the expression

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}$

$=-\frac{2}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}$ ∵ $\frac{-2}{3}\times \frac{3}{5}=-\frac{2}{5}$

$=-\frac{2}{5}+\frac{1}{4}$ ∵ $\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=\frac{1}{4}$

$\mathrm{Least\:Common\:Multiplier\:of\:}5,\:4:\quad 20$

$=-\frac{8}{20}+\frac{5}{20}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-8+5}{20}$

$\mathrm{Add/Subtract\:the\:numbers:}\:-8+5=-3$

$=\frac{-3}{20}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{3}{20}$

Therefore,

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}$

Question # 2

Answer:

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}$

Step-by-step explanation:

Given

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}$

$=-\frac{6}{35}-\frac{1}{6}\times \frac{3}{2}\times \frac{2}{5}\times \frac{1}{14}$ ∵ $\frac{2}{5}\times \frac{-3}{7}=-\frac{6}{35}$

$=-\frac{6}{35}-\frac{1}{140}$ ∵ $\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=\frac{1}{140}$

$\mathrm{Least\:Common\:Multiplier\:of\:}35,\:140:\quad 140$

$=-\frac{24}{140}-\frac{1}{140}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-24-1}{140}$

$\mathrm{Subtract\:the\:numbers:}\:-24-1=-25$

$=\frac{-25}{140}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{25}{140}$

$\mathrm{Cancel\:the\:common\:factor:}\:5$

$=-\frac{5}{28}$

Therefore,

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}$

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2 years ago

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Sophie [7]

Answer:

60ft

Step-by-step explanation:

h x 20 = 120

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2 years ago