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1 year ago

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Mathematics

2 answers:

Tema [17]1 year ago

7 0

Answer: 88

Step-by-step explanation:

Oduvanchick [21]1 year ago

6 0

Answer:

see attached

Step-by-step explanation:

The Pythagorean theorem can be used to find the hypotenuse associated with each pair of legs. That tells you ...

c² = a² +b² . . . . . legs a, b; hypotenuse c

<h3>alternate form of Pythagorean theorem</h3>

For the purpose of this problem, it is convenient to consider a slightly different form of the equation.

For legs √a and √b, the hypotenuse √c is given by ...

(√c)² = (√a)² +(√b)²

c = a +b

That is ...

legs √a, √b ⇒ hypotenuse √(a+b)

<h3>application to this problem</h3>

Since the legs are (mostly) given in terms of square roots, the value under the radical for the hypotenuse is simply the sum of those:

legs: √1, √2 ⇒ hypotenuse √(1+2) = √3

legs: √2, √3 ⇒ hypotenuse √(2+3) = √5

legs: √5, √3 ⇒ hypotenuse √(5+3) = √8

legs: √5, √1 ⇒ hypotenuse √(5+1) = √6

_____

<em>Additional comment</em>

You may not see the leg lengths given as square roots very often. This is a rather unusual set of problems for hypotenuse length.

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Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$