Help on 7 and 8b please and thank you.

Mathematics

1 answer:

andreev551 [17]4 years ago

8 0

7)
$total \: increase \: for \:sides \: of \: the \: equillateral \: triangle = 3 \times 5cm = 15cm \\ perimeter \: of \: an \: equillateral \: triangle = 3 \times length \: of \: one \: side \\ originl \: perimeter = incresed \: perimeter - increase = 60cm - 15cm = 45cm \\ one \: side \: of \: the \: equillateral \: triangle = 45cm \div 3 = 15cm$
8b)
$let \: x \: represent \: the \: amount \: each \: person \: has \: in \: the \: end. \\ 40 - 4(7.50) = 4x \\ 40 - 30 = 4x \\ 10 = 4x \\ 4x = 10 \\ x = 10 \div 4 \\ x = 2.5 \\ each \: boy \: has \: 2.50 \: in \: the \: end$

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mixer [17]

Answer:

7.8 cm and 25.3 cm

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, that is

?² + 7² = 10.5²

?² + 49 = 110.25 ( subtract 49 from both sides )

?² = 61.25 ( take the square root of both sides )

? = $\sqrt{61.25}$ ≈ 7.8 cm ( to 1 dec. place )