Everyone Pplease help with maths problems?

Question

hey everyone, i would be so greatful if someone could answer the following questions for me, i have tired them over and over, and im still not understand it!! very dissapointing i must say, kids from school, wont help, they always keep to themselfs. anyways here are the questions, thanks in advance, any help would be appreciated.

Question 16) Find
a) the third term in (1-2x)^7
b) the fourth term in (1-x/2)^9
c) the fifth term in (1+x^2)^10
d) the sixth term in (3-2x)^8
e) the seventh term in (1-x/2)^10

Question 18) The fifth and sixth rows of Pascals triangle are shown below.

1 4 6 4 1
1 5 10 10 5 1

find the coefficient of x^3 in the expansion of (x+2)^5

Question 19) find the coefficient of x^6 in the expansion of (x^2+2/x)^6

Question 20) Let a be the coefficient of x^2 and b be the coeficient of x the expansion of the polynomial (2x-1)^4. Show that a= -3b

Answer 1

hey wat is this? wat do u study?? i mean which class?

Answer 2

For question 18 those aren't the fifth and sixth rows, those are the four and fifth rows. The number after the first 1 always tells you what row you are on.

I always remember as one power decreases the other increases.

So the terms in order for (x+2)^5 are:
first term = 1(x^5)(2^0),

second term = 5(x^4)(2^1)

third term = 10(x^3)(2^2)

fourth term = 10(x^2)(2^3)

fifth term = 5(x^1)(2^4)

sixth term= 1(x^0)(2^5)

As the x's power declines the 2's power increases (and you always have one more term than the power you are using- because of the 0 exponent).

Since the third term has x^3 in it that's the term you need to simplify:

10(x^3)(2^2)

Square the 2.
10(4)(x^3)

Multiply.
40(x^3)
and the coefficient is 40.


For 19 use the row: 1 6 15 20 15 6 1 (again second number is 6, and notice there will be 7 terms).

(x^2 + 2/x)^6

You may wish to rewrite the second term as 2x^-1 if you are comfortable with negative exponents or leave it alone.

Same basic exercise as before just make sure you distribute the powers carefully when expanding.

first = 1(x^2)^6(2/x)^0 = 1(x^12)

second = 6(x^2)^5(2/x)^1 = 6(x^10)(2/x) = 12x^9

third = 15(x^2)^4(2/x)^2 = 15(x^8)(4/x^2) = 60x^6

And we can stop here becuase we just found the x^6 term; the coefficient is 60.

Each time you do this make sure you are using the correct row of the triangle (make sure the power matches the number directly after the first 1), and let the first term start with the highest power and descend to 0 and the second term start with 0 and ascend to the highest power (highest power being the power the entire binomial was raised to in the first place).

Answer 3

enable's call the unique extensive form x and the hot one y (a million.5x)(.6) = y including 50% to x makes it a million.5x then multiplying that by skill of 0.6 provides us 60% of that (subtracting 40%) and x - 8 = y the staggering extensive form is 8 decrease than the unique. for the reason that we've 2 equations and the x and y are a similar in the two, we can say that for the reason that (a million.5x)(.6) = y and x - 8 = y (a million.5x)(.6) = y = x - 8 (a million.5x)(.6) = x-8 now we are going to positioned our decimals into fractions to make it distinctly =D (3x/2)(3/5) = x - 8 and sparkling up 9x/10 = x - 8 9x/10 - 10x/10 = -80/10 -x/10 = - 80/10 x = 80 there you circulate!

Answer 4

Question 20)
(2x-1)^4=
1*(2x)^4*(-1)^0+
4*(2x)^3*-1)^1+
6*(2x)^2*(-1)^2+
4*(2x)^1*(-1)^3+
1*(2x)^0*(-1)^4

answer
from
6*(2x)^2*(-1)^2
a=24

from
4*(2x)^1*(-1)^3
b=-8

Hence 24=-3*(-8)
therefore
a=-3b