Algebra Help!! 2 Problems!?

Question

The sum of the squares of two consecutive positive integers is 13. What are the integers? (Let n = the smaller positive number)


A.
The integers are 3 and 4.

B.
The integers are 2 and 3.

C.
The integers are 1 and 2.

D.
The integers are 4 and 5.


The length of a rectangle is 3 cm more than twice the width. If the rectangle's area is 35 cm2, find the dimensions of the rectangle. (Let w = the width.)


A.
The rectangle is 3.2 cm wide and 9.4 cm long.

B.
The rectangle is 3.3 cm wide and 9.6 cm long.


C.
The rectangle is 3.5 cm wide and 10.0 cm long.

D.
The rectangle is 3.8 cm wide and 10.6 cm long.

Answer 1

i forgot that **** like a week after school ended lol

Answer 2

1.
Let the 1st positive integer be n
The 2nd positive integer will be n+1
n^2 + (n+1)^2 = 13
n^2 + (n^2+2n+1) = 13
2n^2 + 2n - 12 = 0
Dividing by 2
n^2 + n - 6 = 0
(n + 3)(n - 2) = 0
Either n+3 = 0
n = -3 (rejected as value of n must be positive)
Or n - 2 = 0
n = 2
So the 1st positive integer will be 2 and the 2nd positive integer will be n + 1 = 2+1 = 3

Answer is 2 and 3 (B)

2.
let the width be w
Area = length*width
Area = (2w + 3) *w
Area = 35
(2w + 3)*w = 35
2w^2 + 3w - 35 = 0
(2w - 7)(w + 5) = 0
Either 2w - 7 = 0
2w = 7
w = 3.5
Or w + 5 = 0
w = -5 (rejected, length cannot be negative)

When w = 3.5, length = 2(3.5)+3 = 10
Answer is 3.5cm wide and 10 cm long. (C)

Answer 3

The first answer has to be B. 2x2=4,3x3=9......4+9=13
The second I think is C. Don't be distracted by the first sentence in the question. All the possible answers are rectangles that are 3cm more than twice the width. Instead find which one has an area of 35cm. 3.5x10=35

Answer 4

1. 2^2 + 3^2 = 4 + 9 + 13. [B]

2. l = 2w + 3. lw = 35.
35/w = 2w + 3
35 = 2w^2 + 3w
(2w -7)(w+5) = 0
w = -5 or 7/2.
the width is 3.5cm and the length is 10 cm. [C]

Answer 5

1) B 2nd 3
2) C 3.5 and 10

Answer 6

First one is B for sure

Answer 7

1. b

2. c

Answer 8

first question is choice b.
second question is choice c.