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Soloha48 [4]
3 years ago
10

When adding two numbers, such as 123 and 423, care is taken to first line them up and then add like digits. How does expanding t

his expression to [(1 × 102) + (2 × 101) + (3 × 100)] + [(4 × 102) + (2 × 101) + (3 × 100)] make the operation more like a polynomial addition problem?
Mathematics
1 answer:
zavuch27 [327]3 years ago
7 0
Think of 10^2, 10^1, and 10^0 as x^2, x^1, and x^0.
When you add polynomials, you can only combine like terms.
When you add expanded numbers, you can only combine like powers of 10.

123 + 423 =

= (100 + 20 + 3) + (400 + 20 + 3)

= (1 * 10^2 + 2 * 10^1 + 3 * 10^0) + (4 * 10^2 + 2 * 10^1 + 3 * 10^0)

= (1 * 10^2 + 4 * 10^2) + (2 * 10^1 + 2 * 10^1) + (3 * 10^0 + 3 * 10^0)

= 5 * 10^2 + 4 * 10^1 + 6 * 10^0

= 500 + 40 + 6

= 546
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16 and 17 please I don’t know if it’s the distance formula
Trava [24]

Answer:

#16 = (0, -10)

#17 = (-18, -3)

Step-by-step explanation:

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A student’s course grade is determined by averaging 4 exams. The student’s grades on the first 3 exams are 85, 87, and 89. What
Nady [450]

Answer:

a). At least 99 marks should be scored by the student to get an A.

b). Student has to score at least 59 and less than 99 grades in the 4th exam to get a B.

Step-by-step explanation:

Student's grades on the first three exams are 85, 87, and 89.

Since student's course grade is to be determined by averaging 4 exams.

So average of 4 exams = \frac{85+87+89+x}{4}

Here x is the grade obtained by the student in 4th exam.

If the average of 4 exams to get an A is at least 90 so the expression will be

\frac{85+87+89+x}{4}\geq 90

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To get a B (Average of at least 80) the expression will be

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Therefore, student has to score at least 59 and less than 99 grades in the 4th exam to get a B.

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