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mixas84 [53]
3 years ago
15

Problem: an old book listed that 72 turkeys were bought for $\$\square.\square\square$ each, at a total cost of $\$\square 67.9\

square.$ what was the cost of each turkey, if each $\square$ stands for a digit that was unreadable in the old book
Mathematics
1 answer:
leonid [27]3 years ago
7 0

Answer:

  72 turkeys were bought at $5.11 each for a total cost of $367.92.

Step-by-step explanation:

Trial and error isn't so difficult for this problem. The final price will be an even number of cents. Without those pennies added, the price per turkey must be below an integer number of cents, but within 1/8 cent of that integer number.

  • 167.90/72 = 2.3319... (too high)
  • 267.90/72 = 3.7208... (too high)
  • 367.90/72 = 5.1097... good candidate
  • 467.90/72 = 6.4986... (too low)
  • 567.90/72 = 7.8875 (too low)
  • 667.90/72 = 9.2763... (too low)

For higher first digits of the total cost, the per-turkey cost is more than 3 digits.

For our "good candidate", the per-turkey cost will be $5.11, so the total bill will be $5.11×72 = $367.92  . . . .  matches the required digits.

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Whats the answer to x^3 y^0 z^-2
Elina [12.6K]

Answer:

\frac{x^3}{z^2}

Step-by-step explanation:

There are no given values, therefore we can't determine a definite answer. However, we are able to simplify the expression.

As we all know that any variable or number to the power of 0 is equal to 1. This means that y^0 = 1.

We also know that a negative power can bring the variable down and make it a fraction. This means that z^-^2=\frac{1}{z^2}

Now that we got these two sorted we are able to write it in a simplified form.

Answer: \frac{x^3}{z^2}

5 0
3 years ago
Can someone please help me with this (BTW ignore my writing)
Jobisdone [24]

ANSWER

On photo

EXPLANATION

Use the quadratic formula since the equation doesn't factorize not simplify by 3

5 0
3 years ago
Mike bought a soft drink for 4 dollars and 9 candy bars. He spent a total of 31 dollars. How much did each candy bar cost ?
Andre45 [30]

Answer:

$3 per candy bar

Step-by-step explanation:

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7 0
3 years ago
Mrs. Smith's class is working on a variety of science experiments. Each
SVETLANKA909090 [29]

Answer:

66

Step-by-step explanation:

If there are  <em>n</em>  students, then the number of pairs is \frac{n(n-1)}{2}.

With 12 students, \frac{12(12-1)}{2} = \frac{132}{2}=66 pairs can be formed.

The reason the formula works is this:  Each of the 12 students can be paired with 11 other students (no student is paired with him/her self).  But counting 12 x 11 = 132  counts each pair <u>twice</u>.  Example: student A can be paired with student B,..., student B can be paired with student A.  The pair was counted two times.

See the attached image that shows pairings of 5 students.  There are

5(5 - 1)/2 = 5(4)/2 = 10 pairs.

8 0
3 years ago
Find the value of the variable.
nalin [4]

Answer:

The variable, y is 11°

Step-by-step explanation:

The given parameters are;

in triangle ΔABC;          {}              in triangle ΔFGH;

Segment \overline {AB} = 14         {}               Segment \overline {FG} = 14

Segment \overline {BC} = 27         {}              Segment \overline {GH} = 19

Segment \overline {AC} = 19         {}               Segment \overline {FH} = 2·y + 5

∡A = 32°                       {}                ∡G = 32°

∡A = ∠BAC which is the angle formed by segments \overline {AB} = 14 and \overline {AC} = 19

Therefore, segment \overline {BC} = 27, is the segment opposite to ∡A = 32°

Similarly, ∡G = ∠FGH which is the angle formed by segments \overline {FG} = 14 and \overline {GH} = 19

Therefore, segment \overline {FH} = 2·y + 5, is the segment opposite to ∡A = 32° and triangle ΔABC ≅ ΔFGH by Side-Angle-Side congruency rule which gives;

\overline {FH} ≅ \overline {BC} by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∴ \overline {FH} = \overline {BC} = 27° y definition of congruency

\overline {FH} = 2·y + 5 = 27° by transitive property

∴ 2·y + 5 = 27°

2·y = 27° - 5° = 22°

y = 22°/2 = 11°

The variable, y = 11°

8 0
3 years ago
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