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marusya05 [52]
3 years ago
13

The angle of elevation from a viewer to the center of a fireworks display is 55°. If the viewer is 75 yards away from where the

fireworks are launched, at what height is the center of the display? Round to the nearest whole yard
Mathematics
1 answer:
notka56 [123]3 years ago
5 0

Answer:

107.1075 yards

Step-by-step explanation:

To solve this problem, we can imagine a triangle, where one of the angles is 55 degrees, the adjacent cathetus is 75 yards, and we want to know the opposite cathetus (the height of the center of the display, "h").

The relation between opposite and adjacent cathetus is the tangent, so:

tan(55) = h/75

tan(55) = 1.4281, so:

h/75 = 1.4281

h = 75*1.4281 = 107.1075 yards

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What’s the total value of 20 quarters, 40 dimes, 30 nickels, and 15 pennies
Inessa [10]

Answer$10 65 cents

Step-by-step explanation:

20/4= $5 40/10= $4  .05 x 30= $1.50 and add those together to get $10.50 then add the 15 pennies which is $10.65

6 0
3 years ago
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Can you solve 27,28,39,30,31,32,33,34?
kozerog [31]

Answer:

27. x^7

28. 4x^13

29. 1/f^4

30. x^5

31. 2x^14

32. (4y^3 x^2)^2

33. (2y/x)^6

Step-by-step explanation:

27. \:  \:  \frac{ {x}^{ - 1} }{ {x}^{ - 8} }  \\  \frac{ {x}^{8} }{ {x}^{1} }  \\  {x}^{8 - 1}   =  {x}^{7} \\

28. \:  \:  \:  \frac{ {52x}^{6} }{ {13x}^{ - 7} }  \\   \frac{ {52x}^{6}  {x}^{7} }{13}  \\  {4x}^{6 + 7}  =  {4x}^{13}

29. \:  \: {f}^{ - 3} ( {f}^{2} )( {f}^{ - 3} ) \\  {f}^{( - 3) +2 + ( - 3) }  \\  {f}^{ - 4}  \\  \frac{1}{ {f}^{4} }  \\

30. \:  \: \frac{ {x}^{ - 4} }{ {x}^{ - 9} } \\   \frac{ {x}^{9} }{ {x}^{4} }  \\  {x}^{9 - 4}  =  {x}^{5}

31. \:  \:  \frac{ {24x}^{6} }{ {12x}^{ - 8} }  \\  \frac{ {24x}^{6} {x}^{8}  }{12}  \\  {2x}^{6 + 8}  =  {2x}^{14}  \\

32. \:  \:  \frac{ {3x}^{2}  {y}^{ - 3} }{ {12x}^{6}  {y}^{3} }  \\  \frac{ {3x}^{2} }{ {12x}^{6} {y}^{3}  {y}^{3}  }  \\  \frac{ {x}^{2 - 6} }{ {4y}^{3 + 3} }  \\  \frac{ {x}^{ - 4} }{ {4y}^{6} }  \\  \frac{1}{ {4y}^{6}  {x}^{4} }  = \frac{1}{({ {4y}^{3} {x}^{2} ) }^{2} }

33. \:  \:  {( {2x}^{3}  {y}^{ - 3}) }^{ - 2}  \\ {2x}^{3 \times  - 2}  {y}^{ - 3 \times  - 2}  \\  {2x}^{ - 6}  {y}^{6}  \\  \frac{ {2y}^{6} }{ {x}^{6} }  \\  {( \frac{2y}{x} )}^{6} \\

4 0
2 years ago
Right △ABC has coordinates A(-7, 3), B(-7, 10), and C(-1, 3). The triangle is reflected over the x-axis and then reflected again
aalyn [17]
X-axis = A (-7,-3) B (-7,-10) C (-1,-3) first they changed to the x-axis which is this. Just so you know on the x-axis the x-axis stay and the y-axis changed which is this.
When they reflect again on the y-axis the verter of A' is
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Hope this help!
7 0
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And the result simplifies to

18x^2y^2\sqrt[3]{3xy^2}

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