+0

# MATHS

0
76
6
+502

Pls help me on this question

question 9 part c

Rauhan  Dec 24, 2017

#2
+91451
+1

Hi Rauhan

9. Ann has some sticks that are all of the same length. She arranges them in squares and has made the following 3 rows of patterns:

Row 1

Row 2

Row 3

She notices that 4 sticks are required to make the single square in the first row,

7 sticks to make 2 squares in the second row

and in the third row she needs 10 sticks to make 3 squares.

(a) Find an expression, in terms of n, for the number of sticks required to make a similar arrangement of n squares in the nth row.

 row (r) 1 2 3 ... n match sticks (m) 4 7 10 3n+1

Ann continues to make squares following the same pattern. She makes 4 squares in the 4th row and so on until she has completed 10 rows.

(b) Find the total number of sticks Ann uses in making these 10 rows.

m(10)=3*10+1=31 match sticks

Ann started with 1750 sticks.

Given that Ann continues the pattern to complete k rows but does not have sufficient sticks to complete the (k + 1)th row,

(c) show that k satisfies (3k – 100)(k + 35) < 0.

The total number of sticks in k rows is

$$4+7+10+.......(3k+1) \qquad \text{Sum of an AP}\\ =\frac{n}{2}(a+L)\\ =\frac{k}{2}(4+3k+1)\\ =\frac{k}{2}(3k+5)\\ now\\ \frac{k}{2}(3k+5)<1750\\ k(3k+5)<3500\\ 3k^2+5k<3500\\ 3k^2+5k-3500<0\\ \qquad 3*-3500=-10500\\ \qquad \text{I need 2 numbers that multiply to -10500 and add to +5}\\ \qquad \text{Those numbers are 105 and -100}\\ 3k^2+105k-100k-3500<0\\ 3k(k+35)-100(k+35)<0\\ (3k-100)(k+35)<0$$

(d) Find the value of k.

if you graph y=(3k-100)(k+35)

it will be a concave up parabola and y will be less then 0  between the two zeros.

3k-100=0

3k=100

k=33 and a 1/3

and

k=35=0

k=-35

k cant be negative so Ann will have enough sticks for 1 to 33 rows.

She will not have enough sticks for 34 rows.

So

k=33

Melody  Dec 24, 2017
Sort:

#1
+202
+1

(3k – 100)(k + 35) < 0.

You want every (...) to have different sign so

the first () if k>33.333  ()>0 if k<33.333 ()<0                κ              -35           33.333

the second () if k>-35 ()>0 if k<-35 ()<0                     (1)      -              -                    +

(2)      -             +                    +

Finally    +              -                    +

You want  (3k – 100)(k + 35) be <0 so       -35

κ  between  -35 and 33.333

Whitespy001  Dec 24, 2017
edited by Whitespy001  Dec 24, 2017
#3
+91451
0

Thanks Whitespy

But... you have not actually answered part c.

Melody  Dec 24, 2017
#4
+202
0

yes because i understand false the question :p Im sorry :(

Whitespy001  Dec 24, 2017
#5
+91451
+1

I understood that, it was not a problem :)

Melody  Dec 24, 2017
#2
+91451
+1

Hi Rauhan

9. Ann has some sticks that are all of the same length. She arranges them in squares and has made the following 3 rows of patterns:

Row 1

Row 2

Row 3

She notices that 4 sticks are required to make the single square in the first row,

7 sticks to make 2 squares in the second row

and in the third row she needs 10 sticks to make 3 squares.

(a) Find an expression, in terms of n, for the number of sticks required to make a similar arrangement of n squares in the nth row.

 row (r) 1 2 3 ... n match sticks (m) 4 7 10 3n+1

Ann continues to make squares following the same pattern. She makes 4 squares in the 4th row and so on until she has completed 10 rows.

(b) Find the total number of sticks Ann uses in making these 10 rows.

m(10)=3*10+1=31 match sticks

Ann started with 1750 sticks.

Given that Ann continues the pattern to complete k rows but does not have sufficient sticks to complete the (k + 1)th row,

(c) show that k satisfies (3k – 100)(k + 35) < 0.

The total number of sticks in k rows is

$$4+7+10+.......(3k+1) \qquad \text{Sum of an AP}\\ =\frac{n}{2}(a+L)\\ =\frac{k}{2}(4+3k+1)\\ =\frac{k}{2}(3k+5)\\ now\\ \frac{k}{2}(3k+5)<1750\\ k(3k+5)<3500\\ 3k^2+5k<3500\\ 3k^2+5k-3500<0\\ \qquad 3*-3500=-10500\\ \qquad \text{I need 2 numbers that multiply to -10500 and add to +5}\\ \qquad \text{Those numbers are 105 and -100}\\ 3k^2+105k-100k-3500<0\\ 3k(k+35)-100(k+35)<0\\ (3k-100)(k+35)<0$$

(d) Find the value of k.

if you graph y=(3k-100)(k+35)

it will be a concave up parabola and y will be less then 0  between the two zeros.

3k-100=0

3k=100

k=33 and a 1/3

and

k=35=0

k=-35

k cant be negative so Ann will have enough sticks for 1 to 33 rows.

She will not have enough sticks for 34 rows.

So

k=33