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kenny6666 [7]
2 years ago
6

50 PTS!!A right triangle has a leg measuring 16in. And 28in.

Mathematics
2 answers:
Leokris [45]2 years ago
7 0

Answer:

2)  32.2 in (nearest tenth)

Step-by-step explanation:

<u>Pythagoras’ Theorem</u>

\sf a^2+b^2=c^2

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

Given:

  • a = 16 in
  • b = 28 in
  • c = hypotenuse

Substituting these values into the formula and solving for c:

\implies \sf 16^2+28^2=c^2

\implies \sf 256+784=c^2

\implies \sf c^2=1040

\implies \sf c=\pm\sqrt{1040}

\implies \sf c=\pm 32.2\:in\:(nearest\:tenth)

As distance is positive, c = 32.2 in (nearest tenth)

andreev551 [17]2 years ago
7 0
  • 32.2 inches. (Option 2)

Step-by-step explanation :

Here, A right angled triangle is given with the measure of two sides and we are to find the measure of the third side.

We'll find the measure of third side with the help of the Pythagorean theorem,

\\ {\longrightarrow \pmb{\sf {\qquad (Hypotenuse {)}^{2}= (Base) {}^{2} + (Perpendicular {)}^{2} }}} \\  \\

Here,

  • The base (BC) is 16 in

  • The perpendicular (AB) is 28 in

  • The hypotenuse is AC.

\:

So, substituting the values in the formula we get :

\\ {\longrightarrow \pmb{\sf {\qquad (AC {)}^{2}= (16) {}^{2} + (28 {)}^{2} }}} \\ \\

{\longrightarrow \pmb{\sf {\qquad (AC {)}^{2}=256 + 784 }}} \\ \\

{\longrightarrow \pmb{\sf {\qquad (AC {)}^{2}=1040 }}} \\ \\

{\longrightarrow \pmb{\sf {\qquad AC = \sqrt{1040}  }}} \\ \\

{\longrightarrow \pmb{\frak {\qquad AC = 32.24  }}} \\ \\

Therefore,

  • The length of the hypotenuse (AC) is 32.2 in (Rounded to nearest tenth)

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Answer: #7 D) neither set. #6. C) the sum of the data values must be 75. #5. The 6th missing temperature must be 80 degrees

Step-by-step explanation: #7. if you eliminate 3 numbers on each side on set A, then the numbers remaining are 2 and 19. In order to find the median add the two numbers, 19+2 which equals 21. Then divide that number by 2 to find your median, or in other words find the average, 21 divided by 2 is 10.5 which is not 12.5, so that set doesn’t work. In set B eliminate 2 numbers on each side. That will leave you with 10 and 9, then again find the average of the numbers. 10+9=19, and 19 divided by 2= 9.5. So therefore neither of the sets have a median of 12.5

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