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riadik2000 [5.3K]
3 years ago
10

Antoine and Tess have a disagreement over how to compute a 15% gratuity on $46.00. Tess says, “It is easy to find 10% of 46 by m

oving the decimal point one place to the left to get $4.60. Do that twice. Then add the two amounts to get $4.60 + $4.60 = $9.20 for the 15% gratuity.”
Mathematics
1 answer:
Nady [450]3 years ago
6 0

Tess is correct about figuring 10% but he added 2 of them together which is like adding 10% +10% which is 20% not 15%.

 he should have taken 1/2 of the 4.60 for 5% since 5 is half of 10

 half of 4.60 = 2.30

then add that to the 4.60

4.60 +2.30 = 6.90 is 15%

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Answer:

cos q = 3/5

Step-by-step explanation:

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2 years ago
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3 years ago
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kumpel [21]

Answer:

  x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

__

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

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The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

  x = -1: √(-1+2) +1 = √(3(-1)+3)   ⇒   1+1 = 0 . . . . not true

  x = 2: √(2+2) +1 = √(3(2) +3)   ⇒   2 +1 = 3 . . . . true . . . x = 2 is the solution

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<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

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The value of x is ...

  x = u² -2 = 2² -2

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<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

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