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Likurg_2 [28]
3 years ago
14

Sharon paid $78 sales tax on a new camera. If the sales tax is 6.5%, what was the cost of the camera?

Mathematics
2 answers:
svlad2 [7]3 years ago
7 0

<em>Answer:</em>

<em>$1,200</em>

<em>Step-by-step explanation:</em>

<em>78 divided by 6.5=12</em>

<em>With percents you have to multiply to get out of 100 so:</em>

<em>12x100=1,200</em>

<em>78 divided by 6.5%=1,200</em>

-<em>SunflowerJada</em>

Irina18 [472]3 years ago
5 0

Answer:

$1200

Step-by-step explanation:

$1200

Step-by-step explanation:

78 = 6.5%

This means that if you divide 78 by 6.5, you get the equivalent of 1% of the price:

78 ÷ 6.5 = 12

So 1% = 12

Now simply multiply this by 100 to get the full answer:

12 x 100 = 1200

So the camera cost $1200!

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100 points! simplify write as a product compute
Rom4ik [11]

Answer:

a) \sqrt{61 - 24 \sqrt{5} }  =  - 4  + 3 \sqrt{5}

b)( \sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) }  + {2 \sqrt{bc}  } )

c) \frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }  =   - 1

Step-by-step explanation:

We want to simplify

\sqrt{61 - 24 \sqrt{5} }

Let :

\sqrt{61 - 24 \sqrt{5} }  = a - b \sqrt{5}

Square both sides of the equation:

(\sqrt{61 - 24 \sqrt{5} } )^{2}  =  ({a - b \sqrt{5} })^{2}

Expand the RHS;

61 - 24 \sqrt{5} =  {a}^{2}  - 2ab \sqrt{5}  + 5 {b}^{2}

Compare coefficients on both sides:

{a}^{2}  + 5 {b}^{2}  = 61 -  -  - (1)

- 24 =  - 2ab \\ ab = 12 \\ b =  \frac{12}{b}  -  -  -( 2)

Solve the equations simultaneously,

\frac{144}{ {b}^{2} }  + 5 {b}^{2}  = 61

5 {b}^{4}  - 61 {b}^{2}  + 144 = 0

Solve the quadratic equation in b²

{b}^{2}  = 9 \: or \:  {b}^{2}  =  \frac{16}{5}

This implies that:

b =  \pm3 \: or \: b =  \pm  \frac{4 \sqrt{5} }{5}

When b=-3,

a =  - 4

Therefore

\sqrt{61 - 24 \sqrt{5} }  =  - 4  + 3 \sqrt{5}

We want to rewrite as a product:

{b}^{2}  {c}^{2}  - 4bc -  {b}^{2}  -  {c}^{2}  + 1

as a product:

We rearrange to get:

{b}^{2}  {c}^{2}   -  {b}^{2}  -  {c}^{2}  + 1- 4bc

We factor to get:

{b}^{2} ( {c}^{2}   -  1)  -  ({c}^{2}   -  1)- 4bc

Factor again to get;

( {c}^{2}   -  1) ({b}^{2}   -  1)- 4bc

We rewrite as difference of two squares:

(\sqrt{( {c}^{2}   -  1) ({b}^{2}   -  1) })^{2} - ( {2 \sqrt{bc} })^{2}

We factor the difference of square further to get;

( \sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) }  + {2 \sqrt{bc}  } )

c) We want to compute:

\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }

Let the numerator,

\sqrt{9 - 4 \sqrt{5} }  = a - b \sqrt{5}

Square both sides of the equation;

9 - 4 \sqrt{5}  =  {a}^{2}  - 2ab \sqrt{5}  + 5 {b}^{2}

Compare coefficients in both equations;

{a}^{2}  + 5 {b}^{2}  = 9 -  -  - (1)

and

- 2ab =  - 4 \\ ab = 2 \\ a =  \frac{2}{b}  -  -  -  - (2)

Put equation (2) in (1) and solve;

\frac{4}{ {b}^{2} }  + 5 {b}^{2}  = 9

5 {b}^{4}   - 9 {b}^{2}  + 4 = 0

b =  \pm1

When b=-1, a=-2

This means that:

\sqrt{9 - 4 \sqrt{5} }  =  - 2 +  \sqrt{5}

This implies that:

\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }  =  \frac{ - 2 +  \sqrt{5} }{2 -  \sqrt{5} }  =  \frac{ - (2 -  \sqrt{5)} }{2 -  \sqrt{5} }  =  - 1

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So, computing this one at a time, and summing it all up

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