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shtirl [24]
3 years ago
12

One leg of an isosceles right triangle measures 5 inches. Rounded to the nearest tenth, what is the approximate length of the hy

potenuse?
Mathematics
2 answers:
Komok [63]3 years ago
5 0
Since it is isosceles right triangle, both legs = 5 inches.

using pythagoras theorem,


hypotenuse^2 = 5^2 +5^2 = 25+25 = 50
taking square root on both sides,
hypotenuse = sqrt 50  = 5 sqrt 2 or 7.071

rounded to nearest tenth,
length of hypotenuse is 7.1 inches.
gavmur [86]3 years ago
3 0

Answer:

the answer is 7.1

Step-by-step explanation:

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Step-by-step explanation:

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5/7 divided by 20/5 <br> then simply it completely
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Answer:

5 / 28

Step-by-step explanation:

5/7 ÷ 20/5

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A projectile is launched upward from a height of 704 feet with an initial velocity of 112 ft/s. The equation gives the height h
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4 years ago
In the figure, triangle CAE is an enlargement of triangle CBD with scale factor of 4/3. The area of the smaller triangle is 9cm2
balandron [24]

Answer:

16 cm^2

Step-by-step explanation:

Given

\triangle CAE -- Bigger Triangle

\triangle CBD -- Smaller Triangle

k = \frac{4}{3} --- Scale factor

Area of CBD = 9

Required

Determine the area of CAE

The area of triangle CBD is:

A_1 = \frac{1}{2}bh

\frac{1}{2}bh = 9

The area of CAE is:

A_2 = \frac{1}{2}BH

Where:

B = \frac{4}{3}b and

H = \frac{4}{3}h

The above values is the dimension of the larger triangle (after dilation).

So, we have:

A_2 = \frac{1}{2}*\frac{4}{3}b * \frac{4}{3} * h

A_2 = \frac{1}{2}*\frac{4}{3} * \frac{4}{3} *b* h

A_2 = \frac{1}{2}*\frac{16}{9}  *b* h

Re-order

A_2 = \frac{16}{9}*\frac{1}{2}* b* h

A_2 = \frac{16}{9}*\frac{1}{2}bh

Recall that:

\frac{1}{2}bh = 9

A_2 = \frac{16}{9}*9

A_2 = 16

Hence, the area is 16 cm^2

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3 years ago
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