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vampirchik [111]
1 year ago
7

7. Use the given information to find the total cost. Do not estimate other than to round to the nearest cent, if necessary. Food

bill: $68.55 Sales Tax: 6%
Mathematics
1 answer:
11111nata11111 [884]1 year ago
7 0

Input data

Food bill = $68.55

Tax = 6%

Procedure

T: Total cost

\begin{gathered} T=68.55+68.55\cdot6\text{ \%} \\ T=68.55+4.113 \\ T=72.663 \end{gathered}

The total cost would be $ 72.66

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Answer:

If 7 minus 5 equals 2, then 7 minus 2 equals 5.

Step-by-step explanation:

This is because of the associative property of equality.

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3 years ago
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Maru [420]

Answer:

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Step-by-step explanation:

To calculate the geometric mean we need to first of all multiply 24 and 32 and take the square root of it (i.e. 24*32 is 768, √768 is 27.712.....). However, in this case, we need to represent the answer in a simplified surd. To do this we need to find the highest possible perfect square that is below 768. Here it is 27 because 27*27 equals 729. Now we can go ahead and subtract 768 by 729. We get 39. So now we got two different surds. √729 and √39. We can simplify the √729 to 27. Thus our answer is the combination of both 27*√39 or 27√39.

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Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
GarryVolchara [31]

<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

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4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

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<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

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<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \overline{AD} \cong \overline{DB} (definition of a bisector)

5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

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