What is the length of side x,y, and z

Mathematics

1 answer:

Bezzdna [24]2 years ago

4 0

Answer:

X = $\sqrt{2}$ , Y = 3 , Z = 10

Step-by-step explanation:

using Pythagoras' identity in all 3 right triangles

X² = 1² + 1² = 1 + 1 = 2 ( take square root of both sides )

X = $\sqrt{2}$

middle triangle

Y² = 1² + X² = 1 + ( $\sqrt{2}$ )² = 1 + 2 = 3

triangle on right

Z² = 1² + Y² = 1 + 3² = 1 + 9 = 10

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the pathway of sams dive into the swimming pool can be represented by the equation h=-16t²+15t+12 (where t=time in seconds and h

aliya0001 [1]

Given the equation representing the pathway of sam's dive into the swimming pool, his height after two seconds is -22ft.

<h3>What is a Quadratic Equation?</h3>

Quadratic equation is simply an algebraic expression of the second degree in x. Quadratic equation in its standard form is;

ax² + bx + c = 0

Where x is the unknown

Given the data in the question

Equation of the pathway of sams dive into the swimming pool is h = -16t² + 15t + 12.
Time t = 2.0 seconds
Height h = ?

Now, to get the height after 2 seconds, we substitute 2 for t in the equation.

h = -16t² + 15t + 12.

If t = 2

h = -16(2)² + 15(2) + 12

h = -16(4) + 30 + 12

h = -64 + 30 + 12

h = -64 + 42

h = -22ft

Therefore, given the equation representing the pathway of sam's dive into the swimming pool, his height after two seconds is -22ft.

Learn more about quadratic equations here: brainly.com/question/1863222

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2 years ago

Find the quotient -506/-8

Vlad1618 [11]

The answer to your question is 16.25.. just use a calculator

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3 years ago

Read 2 more answers

Right triangles and trigonometry help!

Maslowich

Step-by-step explanation:

$sin \: 45 \degree = \frac{39}{z} \\ \\ \therefore \: \frac{1}{ \sqrt{2} } = \frac{39}{z} \\ \\ \therefore \: z = 39 \sqrt{2} \\ let \: p \: be \: the \: perpendicular \: \\ \\\therefore \: tan \: 45 \degree = \frac{39}{p} \\ \\ \therefore \:1 = \frac{39}{p} \\ \\ \therefore \:p = 39 \\ \\ sin \: 60 \degree = \frac{39}{x} \\ \\ \therefore \: \frac{ \sqrt{3} }{2} = \frac{39}{x} \\ \\ \therefore \: x = \frac{78}{ \sqrt{3} } \\ \\ \therefore \: x = \frac{78 \sqrt{3} }{ 3 } \\ \\ \therefore \: x = 26 \sqrt{3} \\\\tan\: 60 \degree = \frac{39}{y} \\ \\ \therefore \: \sqrt{3} = \frac{39}{y} \\ \\ \therefore \: y = \frac{39}{ \sqrt{3} } \\ \\ \therefore \: y= \frac{39 \sqrt{3} }{ 3 } \\ \\ \therefore \: y = 13 \sqrt{3} \\$