Answer:
it is not a function
Step-by-step explanation:
because the element 2 has not ot unique image
4 1/5 + B = 9 3/5
21/5 + B = 48/5
B = 27/5
B = 5 2/5
Letter C is the correct answer!
Hope this helps! ;)
Your answer is A, since there is a variable and two numbers, two numbers is the binomial part. Hope this helped!
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Answer:
(a) 1. Distributive property 2. Combine like terms 3. Addition property of equality 4. Division property of equality
Step-by-step explanation:
Replacement of -1/2(8x +2) by -4x -1 is use of the <em>distributive property</em>, eliminating choices B and D.
In step 3, addition of 1 to both sides of the equation is use of the <em>addition property of equality</em>, eliminating choice C. This leaves only choice A.
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<em>Additional comment</em>
This problem makes a distinction between the addition property of equality and the subtraction property of equality. They are essentially the same property, since addition of +1 is the same as subtraction of -1. The result shown in Step 3 could be from addition of +1 to both sides of the equation, or it could be from subtraction of -1 from both sides of the equation.
In general, you want to add the opposite of the number you don't want. Here, that number is -1, so we add +1. Of course, adding an opposite is the same as subtracting.
In short, you can argue both choices A and C have correct justifications. The only reason to prefer choice A is that we usually think of adding positive numbers as <em>addition</em>, and adding negative numbers as <em>subtraction</em>.
Answer:
The value of Car B will become greater than the value of car A during the fifth year.
Step-by-step explanation:
Note: See the attached excel file for calculation of beginning and ending values of Cars A and B.
In the attached excel file, the following are used:
Annual Depreciation expense of Car A = Initial value of Car A * Depreciates rate of Car A = 30,000 * 20% = 6,000
Annual Depreciation expense of Car B from Year 1 to Year 6 = Initial value of Car B * Depreciates rate of Car B = 20,000 * 15% = 3,000
Annual Depreciation expense of Car B in Year 7 = Beginning value of Car B in Year 7 = 2,000
Conclusion
Since the 8,000 Beginning value of Car B in Year 5 is greater than the 6,000 Beginning value of Car A in Year 5, it therefore implies that the value Car B becomes greater than the value of car A during the fifth year.
