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Marina CMI [18]
2 years ago
12

Show all steps question is in the photo.

Mathematics
2 answers:
Lady bird [3.3K]2 years ago
5 0

\\ \sf\longmapsto \sqrt{108x^2y^3}-\sqrt{12x^2y^3}+x\sqrt{75y^3}

\\ \sf\longmapsto \sqrt{2\times 2\times 3\times 3\times 3\times x\times x\times y\times y\times y}-\sqrt{2\times 2\times 3\times x\times x\times y\times y\times y}+\sqrt{5\times 5\times 3\times y\times y\times y}

\\ \sf\longmapsto 6\sqrt{3}xy\sqrt{y}-2\sqrt{3}xy\sqrt{y}+5\sqrt{3}xy\sqrt{y}

\\ \sf\longmapsto \sqrt{3}xy\sqrt{y}(6-2+5)

\\ \sf\longmapsto 9\sqrt{3}xy\sqrt{y}

belka [17]2 years ago
3 0

Answer:

{ \rm{ \sqrt{108 {x}^{2}  {y}^{3} } -  \sqrt{12 {x}^{2} {y}^{3}  }  + x \sqrt{75 {y}^{3} }  }} \\  \\  = { \rm{\sqrt{(36 \times 3) {x}^{2}  {y}^{3} } -  \sqrt{(4 \times 3) {x}^{2} {y}^{3}  }  + x \sqrt{(3 \times 25)}   {y}^{3} }} \\  \\  = { \rm{ \sqrt{ {x}^{2}  {y}^{3} } ( \sqrt{36 \times 3}  -  \sqrt{4 \times 3}  +  \sqrt{3 \times 25} }}) \\  \\ { =  \rm{x {y}^{ \frac{3}{2} } (6 \sqrt{3} - 2 \sqrt{3}  + 5 \sqrt{3} ) }} \\  \\  = { \rm{x {y}^{ \frac{3}{2} }  \{(6 - 2 + 5) \sqrt{3} \} }} \\  \\  = { \rm{x {y}^{ \frac{3}{2} }  \times 9 \sqrt{3} }} \\  \\  = { \boxed{ \rm{9x \sqrt{3 {y}^{3} } }}}

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juin [17]

Answer:

  y = 5·4^x

Step-by-step explanation:

If you have two points, (x1, y1) and (x2, y2), whose relationship can be described by the exponential function ...

  y = a·b^x

you can find the values of 'a' and 'b' as follows.

Substitute the given points:

  y1 = a·b^(x1)

  y2 = a·b^(x1)

Divide the second equation by the first:

  y2/y1 = ((ab^(x2))/(ab^(x1)) = b^(x2 -x1)

Take the inverse power (root):

  (y2/y1)^(1/(x2 -x1) = b

Use this value of 'b' to find 'a'. Here, we have solved the first equation for 'a'.

  a = y1/(b^(x1))

In summary:

  • b = (y2/y1)^(1/(x2 -x1))
  • a = y1·b^(-x1)

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For the problem at hand, (x1, y1) = (2, 80) and (x2, y2) = (5, 5120).

  b = (5120/80)^(1/(5-2)) = ∛64 = 4

  a = 80·4^(-2) = 80/16 = 5

The exponential function is ...

  y = 5·4^x

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2 years ago
What is the perimeter of this quadrilateral?<br> (5,5)<br> (2, 4)<br> (4, 1)<br> (6, 1)
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BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill CD=\sqrt{[ 6- 4]^2 + [ 1- 1]^2} \\\\\\ CD=\sqrt{2^2+0^2}\implies \boxed{CD=2} \\\\[-0.35em] ~\dotfill\\\\ D(\stackrel{x_1}{6}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) ~\hfill DA=\sqrt{[ 5- 6]^2 + [ 5- 1]^2}

DA=\sqrt{(-1)^2+4^2}\implies \boxed{DA=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{\sqrt{10}~~ + ~~\sqrt{13}~~ + ~~2~~ + ~~\sqrt{17}}~~ \approx ~~ 12.89

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Answer:

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Let y represent the number of student tickets sold.

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x+y≥200x+y≥200

12x+8y≤20012x+8y≤200

x+y≤200x+y≤200

12x+8y≥1000

hope this helps

Step-by-step explanation:

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