1. use pascal's triangle to expand the binomial.
(d – 3)6
d6 – 18d5 + 135d4 – 540d3 + 1,215d2 – 1,458d + 729
d6 + 18d5 + 135d4 + 540d3 + 1,215d2 + 1,458d + 729
d6 – 6d5 + 15d4 – 20d3 + 15d2 – 6d + 1
d6 + 6d5 + 15d4 + 20d3 + 15d2 + 6d + 1
2. use the binomial theorem to expand the binomial. (3v + s)5
s5 – 5s4v + 10s3v2 – 10s2v3 + 5sv4 – v5
s5 + 15s4v + 90s3v2 + 270s2v3 + 405sv4 + 243v5
s5 + 45s4v + 270s3v2 + 810s2v3 + 1,215sv4 + 729v5
s5 + 15s4 + 90s3 + 270s2 + 405s + 243
3. what is the fourth term of (d – 4b)3?
b3
–b3
64b3
–64b3

The correct answers are:

1) d⁶-18d⁵+135d⁴-540d³+1215d²-1458d+729

2) 243v⁵+405v⁴s+270v³s²+90v²s³+15vs⁴+s⁵

3) -64b³

Explanation:

We first write out Pascal's Triangle. We start with 1; below this is a row of two 1's; from this line down, we start with 1 and each successive term is the sum of the two above it. I have attached a picture of Pascal's Triangle.

The coefficients of our expanded polynomial are the numbers in each row of the triangle. The first row represents the coefficients of a binomial to the 0 power; the second row represents the 1st power; the third row, the 2nd power; etc.

Using this, the coefficients of a binomial to the 6th power are: 1, 6, 15, 20, 15, 6, 1.

As we expand our polynomial, our first term will be the first coefficient multiplied by the first term of the binomial to the 6th power and the second term to the 0 power. The second term will be the second coefficient multiplied by the first term of the binomial to the 5th power and the second term to the 1st power; the third term will be the third coefficient multiplied by the first term of the binomial to the 4th power and the second term to the 2nd power; etc. In each term of the expansion, the first term of the binomial loses a power while the second term gains a power. This gives us:

1d⁶(-3)⁰+6(d⁵(-3)¹)+15(d⁴(-3)²)+20(d³(-3)³)+15(d²(-3)⁴)+6(d¹(-3)⁵)+1(d⁰(-3)⁶)

= d⁶-18d⁵+15(9d⁴)+20(-27d³)+15(81d²)+6(-243d)+729

= d⁶-18d⁵+135d⁴-540d³+1215d²-1458d+729

2) The coefficients for the 5th power, based on Pascal's Triangle, would be:

1, 5, 10, 10, 5, 1

This gives us:

1(3v)⁵s⁰+5(3v)⁴s¹+10(3v)³s²+10(3v)²s³+5(3v)¹s⁴+1(3v)⁰s⁵

= 243v⁵+5(81v⁴s)+10(27v³s²)+10(9v²s³)+5(3vs⁴)+s⁵

= 243v⁵+405v⁴s+270v³s²+90v²s³+15vs⁴+s⁵

3) The coefficients of the third power in Pascal's Triangle would be:

1, 3, 3, 1

Since there are 4 terms in this row, the first term will have d³ and (-4b)⁰; in each successive term, d will lose an exponent and (-4b) will gain an exponent. In the fourth term, d will have an exponent of 0 and (-4b) will have an exponent of 3; this makes the fourth term

(-4b)³ = -64b³

step-by-step explanation:

their are no angles. sorry cant .

black lol

step-by-step explanation:

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i can tell you jamal now has 3 cookies xd

tommy has 2.

this made me lol.