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Anton [14]
3 years ago
13

A cook has 11.64 pints of olive oil before the dinner service. There were 5.2 pints of olive oil used during the dinner service.

What amount of olive oil is left after the dinner service, with the correct number of significant digits?
A. 6.44 pints

B. 16.84 pints

C. 6.4 pints

D. 17 pints ​
Mathematics
1 answer:
inn [45]3 years ago
4 0

Answer:

c

Step-by-step explanation:

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Solve for z: -21 = z - (-6 - 2z)
Pepsi [2]

Answer:

z = -9

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
  3. Exponents
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  5. Division
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  7. Subtraction
  • Left to Right

Equality Properties

<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define</u>

-21 = z - (-6 - 2z)

<u>Step 2: Solve for </u><u><em>z</em></u>

  1. Distribute negative:                      -21 = z + 6 + 2z
  2. Combine like terms:                     -21 = 3z + 6
  3. Isolate <em>z</em> term:                               -27 = 3z
  4. Isolate <em>z</em>:                                        -9 = z
  5. Rewrite:                                         z = -9
6 0
3 years ago
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Write the lengths in order from greatest to least<br><br> 12.7 ( 12 7/10) 12 3/5 12 3/4
anygoal [31]
12 3/5, 12 7/10, 12 3/4
8 0
3 years ago
Find the area of the shaded sector of the circle. Leave in terms of pi.
qaws [65]

Answer:

a=10.88\pi

Step-by-step explanation:

using the formula below, the central angle is 20° (bc 360-180-160=20)

radius is 14 (28/2=14)

\frac{20}{360} =\frac{a}{\pi r^2} \\ \frac{1}{18} =\frac{a}{\pi 14^2}

\frac{1}{18} =\frac{a}{196\pi}

a=\frac{1*196\pi}{18} \\a=1*10.88\pi\\a=10.88\pi

3 0
3 years ago
I need help in this problem 8v+w+7-8v+2w
vaieri [72.5K]
By using combine like terms,

<span>8v+w+7-8v+2w

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2w+w+7

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The answer for this problem is 3w+7</span>
5 0
3 years ago
You are testing a machine that scores exam papers and assigns grades. Based on the score achieved the grades are as follows: 1-4
Anni [7]

Answer:

8 test cases

Step-by-step explanation:

Equivalence Partitioning also called as equivalence class partitioning. It is abbreviated as ECP. It is a software testing technique that divides the input test data of the application under test into each partition at least once of equivalent data from which test cases can be derived.

An advantage of this approach is it reduces the time required for performing testing of a software due to less number of test cases.

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