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Nikitich [7]
3 years ago
15

If 3x + y = 12, find a when y = -15

Mathematics
2 answers:
nadezda [96]3 years ago
8 0

Answer:

x= 9

Step-by-step explanation:

3x - 15 = 12

add 15 to both sides

3x = 27

divide both by 3

x= 9

Natasha_Volkova [10]3 years ago
7 0

Answer:

9 is the answer

Step-by-step explanation:

3x+-15=12

you add 15 to 12 which is 27, 3x/27=9

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28.2

Step-by-step explanation:

7.2 + 7.3 + 6.9 + 6.8 = 28.2

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A bag contains 6 black marbles and 4 white marbles. Sally takes out a black marble and does not put it back. What is the probabi
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Answer:

\frac{5}{9}

Step-by-step explanation:

Total\ marbles=10\\\\Black=6\\\\White=4

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Total\ remaining\ marbles=9\\\\Remaining\ Black=5\\\\P(Black)=\frac{Remaining\ Black\ marbles}{Total\ remaining\ marbles}\\\\P(Black)=\frac{5}{9}

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What is the following product? Sqr5x^8y^2 sqr10x^3 sqr12y
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Step-by-step explanation:

\sqrt{5 {x}^{8} {y}^{2}  }  \times  \sqrt{10 {x}^{3} } \times  \sqrt{12y}   \\  \\  =   {x}^{4} y \sqrt{5}  \times x \sqrt{10x}  \times 2 \sqrt{3y}  \\  \\  = 2 {x}^{5} y \sqrt{5 \times 10x \times 3y}  \\  \\  =  2 {x}^{5} y \sqrt{5 \times 5 \times 2x \times 3y}  \\  \\ =  2 {x}^{5} y  \times 5\sqrt{ 2x \times 3y}   \\  \\  = 10 {x}^{5} y \sqrt{6xy}  \\

4 0
4 years ago
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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

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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.

Download xlsx
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