year 10 and 11 maths?

Question

S=n ÷2(a+l)
To make n the subject
To make a the subject
To make l the subject

Transpose the formula v²= u² + 2as
To make v the subject
To make u the subject
To make s the subject


Write what you think will be the next two terms in the pattern each time.
Help me to insert the last four numbers
1,3,7,15,.. ,……, ……,…….,
4,9,16,25, ,.. ,……, ……,…….,
7, 8, 10, 14, 22, ,.. ,……, ……,…….,
1, -8, -17, -26, ,.. ,……, ……,…….,


Help me to insert the last only last number
10, 14,18,22,26 and the next one is?
3,4, 6, 10, 18, and the next one is?


Write down the rule fot the following arithmetic sequences
a. {1, 6, 11,…. }
b. {11, 2, -7, ….}
c. {1,2,4,8….}

Answer 1

S=(n)/ 2(a +1)
To make a the subject, move all other variables to the other side.
First, multiply by 2(a + l) to get it out of the denomenator. So we have
2(a + l)S=n
Now, divide by 2S, to get those terms to the other side. So we have
(a + l)= n/(2S)
Now, subtract l to get it to the other side. So we have
a=n/(2s) -l
Ok, now to make n the subject, if you look up a few lines, we had all other variables on the other side and n by itself.
2(a + l)S=n
Ok, now to make l the subject, we can use one of our earlier equations
(a + l)= n/(2S)
Now, instead of subtracting l to the other side, we subtract a. So we have
l = n/(2S) -a

Transpose the formula v^2 = u^2 + 2as
To make v the subject, simply take the square root of both sides(sqrt). So, we have
sqrt(v^2) = sqrt(u^2 + 2as)
Now, sqrt(v^2) = v. So we have
v = sqrt(u^2 + 2as)
To make u the subject, subtract 2as from the right side. So we have
v^2-2as=u^2
Now take the sqrt of both sides
sqrt(v^2 - 2as)= sqrt(u^2)
Similarly, sqrt(u^2) = u. So we have
u = sqrt(v^2 - 2as)
To make s the subject, subtract u^2 from the right side. So we have
v^2 - u^2 = 2as
Now, divide by 2a. So we have
(v^2 - u^2)/(2a) = s

1, 3, 7, 15, ......
Ok, we see that first we have 1,3, 7, 15. That's a change of +2, +4, +8. So, we see that 2^n will be a factor since 2^1=2, 2^2=4 and 2^3=8.
So, let a(0)=1. Then, a(n)=a(n-1) + 2^(n), where a(n) is the nth term and a(n-1) is the previous term. So, for the term after 15, we will have
a(4)=15 + 2^4= 15 + 16= 31. Then, the next term will be
a(5) = 31 + 2^5= 31 + 32= 63. Then, the next term will be
a(6) = 63 + 2^6= 63 + 64= 127. Then, the next term will be
a(7) = 127 + 2^7= 127 + 128 = 255.

4, 9, 16, 25
If we look closely, we notice that these numbers are squares. 4 = 2^2, 9 =3^2, 16=4^2 and 25=5^2. We then see that the number being squares is the previous number + 1. So, we can write an equation for the nth term,a(n), as (n+1)^2. So,
a(1)=(1+1)^2=(2)^2=4
a(2)=(2+1)^2=(3)^2=9
.....
a(5)=(5+1)^2=(6)^2=36
a(6)=(6+1)^2=(7)^2=49
a(7)=(7+1)^2=(8)^2=64
a(8)=(8+1)^2=(9)^2=81

7,8,10,14,22
Again, we see a difference of +1,+2,+4,+8. So, we see 2^n will be a factor since 2^0=1, 2^1=2, 2^2=4, and 2^3=8. Now, we see the first term, a(0), =7.
We want to write this using 2^n, in this case 2^0. We can write this as
a(0) = 6 + 2^n = 6 + 2^(0) = 6 +1 = 7.
Now, try this for the next term.
a(1) = 6 + 2^(1) = 6 + 2= 8. YAY!!!!!
Now, try this for the next term.
a(2) = 6 + 2^(2)= 6 + 4 = 10. YAY!!!!
You can try it for the next two and you see it works for them as well. So, we can describe the nth term as
a(n)=6 + 2^(n)
Thus, the next 4 terms will be
a(5) = 6 + 2^(5)= 6 + 32= 38
a(6) = 6 + 2^(6)= 6 + 64= 70
a(7) = 6+2^(7) = 6 + 128 = 134
a(8)= 6 + 2^(8)= 6 + 256 = 262

1, -8, -17, -26
So, we see a change of -9,-9, and -9. So, we know that -9n will be a factor in this one. Let a(n)= 1-9n and a(0) be the first term. Lets see if this works.
a(0)= 1- 9(0) = 1 - 0 = 1. YAY!!!!!
a(1)= 1-9(1) = 1-9 = -8. YAY!!!!!
We see that this works for the other two as well. So, a(n) = 1-9n. So, the next 4
a(4)=1-9(4)=1-36=-35
a(5)=1-9(5)=1-45=-44
a(6)=1-9(6)=1-54=-53
a(7)=1-9(7)=1-63=-62
I'M OUT OF ROOM!!!!! =( for the next two questions we see a change of +4,+4,+4... so we can write this as 6+4n, where a(1) is the first term. and then +1,+2,+4,+8, so we can use 2^n again and write this as 3+2^n, where a(0) is the first term.
For a, we see a change of +5,+5. So we can write the nth term as 1+5n where
a(0) is the first term. For b, we see a change of -9,-9,-9. So we can write the nth term as 11-9n, where a(0) is the first term. For c, we see the change of 1,2,4 yet AGAIN!!!, so we can use 2^n. Now, THIS time, 2^n describes the whole series when a(0) is the first term(2^0=1, 2^1=2, 2^2=4, 2^3=8<<<<
i hope my thorough explanation helps.

Answer 2

I'll answer a few, dont even know what ur talking bout for the first bit

1,3,7,15,.. ,……, ……,…….,
3-1=2 | 7-3=4 | 15-7=8..... --> 2^x - 1 <--
next four numbers: 31, 63, 127, 255

4,9,16,25, ,.. ,……, ……,…….,
9-4=5 | 16-9=7 | 25-16=9..... --> (x+1)^2 <--
next four numbers: 36, 49, 64, 81

7, 8, 10, 14, 22, ,.. ,……, ……,…….,
8-7=1 | 10-8=2 | 14-10=4 | 22-14=8....... --> 2^(x-1) + 6 <--
next four numbers: 38, 70, 134, 250

The rest of the sequences u can do urself...safe

Answer 3

hi, Q1. Factorize quadratic expressions of the style ax^2 + bx + c. ax² + bx + c = 0 (Factorise a) a [ x² + bx/a + c/a ] = 0 a [ x² + bx/a + (b²/4a² - b²/4a²) + c/a ] = 0 a [ (x² + bx/a + b²/4a²) - b²/4a² + c/a ] = 0 a [ (x + b/2a)² - b²/4a² + c/a ] = 0 a [ (x + b/2a)² - b²/4a² + 4ac/4a² ] = 0 a [ (x + b/2a)² + (4ac - b²)/4a² ] = 0 a(x + b/2a)² + (4ac - b²)/4a = 0 Q2. perceive and factorize quadratic equations that are ideal squares. a² + b² = (a + b)² - 2ab Q.3 perceive and factorize quadratic equations that are the version of two squares. a² - b² = (a + b)(a - b) Q.4 sparkling up quadratic equations of the style ax^2 + bx + c = 0 via factorizing. ax² + bx + c = 0 (Factorise a) a [ x² + bx/a + c/a ] = 0 a [ x² + bx/a + (b²/4a² - b²/4a²) + c/a ] = 0 a [ (x² + bx/a + b²/4a²) - b²/4a² + c/a ] = 0 a [ (x + b/2a)² - b²/4a² + c/a ] = 0 a [ (x + b/2a)² - b²/4a² + 4ac/4a² ] = 0 a [ (x + b/2a)² + (4ac - b²)/4a² ] = 0 a(x + b/2a)² + (4ac - b²)/4a = 0 a(x + b/2a)² = -(4ac - b²)/4a a(x + b/2a)² = (b² - 4ac)/4a (x + b/2a)² = (b² - 4ac)/4a² (x + b/2a) = ± ?[ (b² - 4ac)/4a² ] (x + b/2a) = [ ± ?(b² - 4ac) ] / (2a) x = -(b/2a) ± [ ?(b² - 4ac) ] / (2a) x = [ -b ± ?(b² - 4ac) ]/(2a) wish this facilitates! J

Answer 4

Q1
n=2s(a+l)
a=[n/(2s)] - l
l=[n/(2s)] -a

Q2
V=√(Usquared + 2as)
U=√(Vsquared - 2as)
s=(Vsquared - Usquared)/2a

Q3

a) ...,15,31,63,127,255
b)...., 25,36,49,64,81
c) ...22,38,70,134, 262
d) ... -26,-35,-44,-53,-62

Q4

a) 30
b)34

Q5

a)16
b)-16
c)16