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3 years ago

Hue is 56 in. tall. His friend is 42 in. tall. Hue’s shadow is 24 in. long. How long is his friend’s shadow at the same time?

Mathematics

2 answers:

pashok25 [27]3 years ago

6 0

The answer would be 18 in.

qwelly [4]3 years ago

3 0

42 in is 3/4th of 56 in. 
3/4th of 24 in. is 18 in. 

The length of his friend's shadow at the same time is 
b.) 18 in.

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3 years ago

Read 2 more answers

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dolphi86 [110]

Answer:

Below in bold.

Step-by-step explanation:

c) 3 x (9^2)^3/4 x ((81^3)^5/6

= 3 x 81^3/4 x 81^15/6

= 3 x 81^(3/4 + 15/6)

= 3 x 81^13/4

= 3 x 3^13

= 3^14

= 4,782,969.

f) (5x^-1y^2)^-2 / (25 x^2 y - 1)^2

= 5^-2 x^2y^-4 / 625 x^4y^-2

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= 5^-6 x^-2y^-2

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5 0

3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

7 0

3 years ago