l need help (desperately) wid these maths sums.......plz help???

Question

1. Find all the common zeros of the polynomials y3 + 5y2 - 9y -45 and y3 + 8y2 +15y?
[numbers written after the variable 'y' are exponents ;i.e. 2 for square and 3 for cube]

2. Determine 'k' so that k+2 ,4k-6 and 3k-2 are three consecutive terms of an arithmetic progression?

3. lf H , C and V are respectively the height ,the curved surface area and the volume of a cone ,prove that :-
3(pi)VH3 [cube] -C2H2 [square] + 9v2 [square] = 0 ?

4. Given that one root of the quadratic equation ax2 + bx + c = 0 ,is three times the other root ,show that 3b2 = 16ac ?

5. The areas of three adjacent faces of a cuboid are x,y,z. lf the volume is v ,prove that v2 = xyz ?

thank u very much to all of u who have spent their precious time 2 answer my queries !!!


plz solve few more problems of mine..........plz

6. The base radii of 2 right circular cones of the same height are in the ratio 3:5.Find the ratio of their volumes?

7. Evaluate-
sin^2 5' + sin^ 10' + p +sin^ 85' + sin^ 90' =19/2 ? Find p?
[ ' symbol after the numeric values stands for degrees

thank u very much to all of u who have spent their precious time 2 answer my queries !!!

plz solve few more problems of mine..........plz

6. The base radii of 2 right circular cones of the same height are in the ratio 3:5.Find the ratio of their volumes?

7. Evaluate-
sin^2 5' + sin^ 10' + p +sin^ 85' + sin^ 90' =19/2 ? Find p?
[ ' symbol after the numeric values stands for degrees]

Answer 1

I'm gonna start by answering Q2 and then I'll get the rest a bit later.

2. We know k+2+x=4k-6 (EQ1) (Where x is the difference of each term in the AP)
and 4k-6+x=3k-2 (EQ2)

Taking (EQ1) we get x=3k-8

Substituting we get 4k-6+3k-8=3k-2
Or 4k=12

Therefore, k=3 (Also, x=1 but we don't need to know that)

Next, Q4

Now, the larger root of a quadratic equation is (-b+SQRT(b^2-4ac))/2a and the smaller is (-b-SQRT(b^2-4ac))/2a

Therefore, (-b+SQRT(b^2-4ac))/2a=3(-b-SQRT(b^2-4ac))/2a
Or -b+SQRT(b^2-4ac)=-3b-3SQRT(b^2-4ac)
2b=-4SQRT(b^2-4ac)
b=-2SQRT(b^2-4ac)
b^2=4b^2-16ac
3b^2=16ac, which is the desired result.

Q5. Let the lengths of the edges be a,b and c

Now, ab=x,bc=y and ac=z, also V=abc

So V^2=(a^2)(b^2)(c^2)

Also, xyz=ab*bc*ac=(a^2)(b^2)(c^2)

Therefore, V^2=xyz, which is the desired result.

I'm sorry I have to go now so I can't answer the other 2 but I've picked the 3 that I could do in the time allowed.

Please pick me for best answer!!!

Answer 2

(1) Common roots of two polynoms are also roots of their difference

3y^2 +24 y +45 = 0 y=(-24 +-6)/6 = -3 and -5 .
Now you must test each if???? In this case as the second polyn has no indepndent term you can write it as
y(y^2+8y+15) with roots 0 ( not common -3 and
-5) Test them in the first
(2) The differnce between two consecutive terms is the ratio

so 4k-6 -k-2 =r and
3k-2-4k+6=r
3k-8=r
-k+4 =r
substracting 4k-12 =0 k=3 r= 1
Now carry on

Answer 3

1)at common zeros:
y3 + 5y2 - 9y -45 = y3 + 8y2 +15y
=> 0=3y2+24y+45
0=y2+8y+15
0=(y+5)(y+3)

y=-3 or -5

Answer 4

Answer to number 2
k=3
k+2=5
4k-6=6
3k-2=7
5,6 and 7 are consecutive numbers

Answer 5

try to understand that : i = sqrt(-a million) so, i^2= -a million i^3= (i^2)i = (-a million)i = -i i^4= (i^2)(i^2) = (-a million)(-a million) = a million i^5= (i^4)i = a million*i = i ... then 7i^6 = 7(i^2)^3 = 7(-a million)^3 = -7 10i^23 = 10(i^22)i = 10((i^2)^11)i = 10((-a million)^11)i = 10(-a million)i = -10i 4i^40 = 4(i^4)^10 = 4(a million)^10 = 4 3i = 3i and, 7i^6+10i^23-4i^40-3i = (-7)+(-10i)-4-3i = -11-13i