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3 years ago

If a 45º-45º-90º triangle has a hypotenuse of length 9, find the leg length.

Mathematics

1 answer:

iVinArrow [24]3 years ago

7 0

Answer:

Step-by-step explanation:

x²+x²=9²

2x²=81

x²=81/2=(81×2)/4

x=9√2/2≈4.5√2≈6.36

x²+x²=11²

2x²=11²

x²=11²/2

x=11/√2=11√2/2≈7.78

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2 years ago

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3 years ago

a class with 30 students had an average score of 80 on a test. A class with 20 students had an average score of 90 on the same t

snow_tiger [21]

The average score of all the students in both classes is 84.

<h3>How to calculate the average?</h3>

The class with 30 students had an average score of 80 on a test. The total score will be:

= 30 × 80

= 2400

A class with 20 students had an average score of 90 on the same test. The total score will be:

= 20 × 90

= 1800

Total scores = 2400 + 1800 = 4200

Number of students = 20 + 30 = 50

The average score will be:

= Total score / Total students

= 4200 / 50

= 84

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8 0

1 year ago

Find the value of x in the figure

liubo4ka [24]

Answer: Therefore it must be the first one, 40

Step-by-step explanation: Vertical angles are always congruent, which means that they are equal. x = 40

8 0

3 years ago

The grade appeal process at a university requires that a jury be structured by selecting six individuals randomly from a pool of

romanna [79]

Answer: a) $\dfrac{3}{874}$

b) $\dfrac{30}{3059}$

c) $\dfrac{3575}{12236}$

Step-by-step explanation:

Given : Number of students : 11

Number of faculty members : 13

Total persons : $11+13=24$

Total number of ways to structure a jury of six people from a group of 24 people :-

$^{24}C_{6}=\dfrac{24!}{6!(24-6)!}=134596$

a) Number of ways of selecting a jury of all students :-

$^{11}C_{6}=\dfrac{11!}{6!(11-6)!}=462$

Then , the probability of selecting a jury of all students :-

$\dfrac{462}{134596}=\dfrac{3}{874}$

b) Number of ways of selecting a jury of all faculty :-

$^{13}C_{6}=\dfrac{13!}{6!(13-6)!}=462$

Then , the probability of selecting a jury of all students :-

$\dfrac{1716}{134596}=\dfrac{30}{3059}$

c) Number of ways of selecting a jury of two students and four faculty :-

$^{11}C_{2}\times^{13}C_{4}=\dfrac{11!}{2!(11-2)!}\times\dfrac{13!}{4!(13-4)!}=39325$

Then , the probability of selecting a jury of all students :-

$\dfrac{39325}{134596}=\dfrac{3575}{12236}$

3 0

3 years ago