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3 years ago

find the approximate length of the leg of a right triangle with one leg length 8 and hypotenuse length 19

Mathematics

2 answers:

Trava [24]3 years ago

8 0

Pythagorean :

hypotenuse^2= base^2+height ^2
19^2=8^2+height ^2
height^2=(19-8)(19+8)
height ^2=11.27
height ~17,74

Anika [276]3 years ago

3 0

So, teh exact value is sqrt( 19^2 - 8^2 ) = sqrt( 361-64) = sqrt(297).

There's people who can calculate this mentally, but I can't. Approximate by known squares: 15^2 = 225, 16^2 = 256, 17^2= 289, 18^2=324,

So, 18.

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295 feet = _____ meters ?

katen-ka-za [31]

Answer:

Hello! I hope I am correct!

Step-by-step explanation:

295 feet will equal to 89.916 meters.

Steps in order to solve this problem:

First you have to multiply the length in feet by 0.3048.

295x 0.3048 or 295* 0.3048

(295 feet is equivalent to 89.916 meters!)

Will equal 89.916 meters.

Therefore, 295 feet = 89.916 meter

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3 years ago

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omeli [17]

Answer:

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2 years ago

If (6y − 11)(6y 11) = ay^2 − b, what is the value of a? the value of a is ____________.

Triss [41]

(6y - 11)(6y + 11) = ay² - b |use (a - b)(a + b) = a² - b²

(6y)² - 11² =ay² - b

36y² - 121 = ay² - b |add b to both sides

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a = (36y² - 121 + b)/y²

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7 0

2 years ago

Solve for the roots in simplest form by completing the square <br><br> x^2-12x+20=0

Alex17521 [72]

Answer:

x = 10 or x = 2

Step-by-step explanation:

Solve for x:

x^2 - 12 x + 20 = 0

Hint: | Solve the quadratic equation by completing the square.

Subtract 20 from both sides:

x^2 - 12 x = -20

Hint: | Take one half of the coefficient of x and square it, then add it to both sides.

Add 36 to both sides:

x^2 - 12 x + 36 = 16

Hint: | Factor the left hand side.

Write the left hand side as a square:

(x - 6)^2 = 16

Hint: | Eliminate the exponent on the left hand side.

Take the square root of both sides:

x - 6 = 4 or x - 6 = -4

Hint: | Look at the first equation: Solve for x.

Add 6 to both sides:

x = 10 or x - 6 = -4

Hint: | Look at the second equation: Solve for x.

Add 6 to both sides:

Answer: x = 10 or x = 2

5 0

2 years ago

Henry ran five races, each of which was a different positive integer number of

dolphi86 [110]

Answer:

10 miles.

Step-by-step explanation:

Let x be the number of miles on Henry's longest race.

We have been given that Henry ran five races, each of which was a different positive integer number of miles.

We can set an equation for the average of races as:

$\frac{\text{The sum distances of 5 races}}{5}=4$

As distance covered in each race is a different positive integer, so let his first four races be 1, 2, 3, 4.

Now let us substitute the distances of 5 races as:

$\frac{1+2+3+4+x}{5}=4$

$\frac{10+x}{5}=4$

Let us multiply both sides of our equation by 5.

$\frac{10+x}{5}*5=4*5$

$10+x=20$

Let us subtract 10 from both sides of our equation.

$10-10+x=20-10$

$x=10$

Therefore, the maximum possible distance of Henry's longest race is 10 miles.

7 0

2 years ago