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Evgesh-ka [11]
3 years ago
13

What is the length of BC in the right triangle below

Mathematics
2 answers:
alex41 [277]3 years ago
8 0
The answer would be 15 units
Anna [14]3 years ago
6 0

Answer:

<h2><em>1</em><em>5</em><em> </em><em>units</em></h2>

<em>The </em><em>length </em><em>of </em><em>BC </em><em>is </em><em>1</em><em>5</em><em> </em><em>units.</em>

<em>Solution,</em>

<em>Hypotenuse(</em><em>h)</em><em>=</em><em>?</em>

<em>perpendicular(</em><em>p)</em><em>=</em><em>1</em><em>2</em>

<em>base(</em><em>b)</em><em>=</em><em>9</em>

<em>Using </em><em>Pythagoras</em><em> </em><em>theorem</em><em>,</em>

<em>{h}^{2}  =  {p}^{2}  +  {b}^{2}  \\ or \:  {h}^{2}  =  {12}^{2}  +  {9}^{2}  \\ or \:  {h}^{2}  = 12  \times 12 + 9  \times 9 \\ or \:  {h}^{2}  = 144 + 81 \\ or \:  {h}^{2}  = 225 \\ or \: h =  \sqrt{225}  \\ or \: h =  \sqrt{ {15}^{2} }  \\ h = 15</em>

<em>hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em>

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A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40
malfutka [58]
PART A

The given equation is

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In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

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We add and subtract

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to get,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4

We again factor -16 out of the first two terms to get,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4

This implies that,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4

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The vertex of this function is

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29

PART B

When the ball hits the ground,

h(t) = 0

This implies that,

- 16 ( t- \frac{5}{4}) ^{2} +29 = 0

We add -29 to both sides to get,

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( t- \frac{5}{4}) ^{2} = \frac{29}{16}

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t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}

t = \frac{ 5 + \sqrt{29} }{4} = 2.60

or

t = \frac{ 5 - \sqrt{29} }{4} = - 0.10

Since time cannot be negative, we discard the negative value and pick,

t = 2.60s
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