The square of a number exceeds that number by 12. What are the two possible solutions?

Mathematics

2 answers:

aev [14]3 years ago

7 0

Answer:

The two possible solutions are -3 or 4.

Step-by-step explanation:

$x^{2}=x+12\\x^{2} -x-12=0\\x^{2}-+3x-4x-12=0\\x(x+3)-4(x+3)=0\\(x-4)(x+3)=0\\x=-4/x=3$

laiz [17]3 years ago

4 0

Answer:

The correct answer is option C.

Step-by-step explanation:

Let the number be x.

The square of a number exceeds that number by 12 .Mathematically can be written as:

$x^2-x=12$

$x^2-x-12=0$

Using middle term splitting to factorize the expression:

$x^2-4x+3x-12=0$

$x(x-4)+(3)(x-4)=0$

$(x+3)(x-4)=0$

The two possible solutions are :

1) (x+3)=0

x = -3

2) (x-4)=0

x = 4

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70 decreased by 13% =

Pie

If you have a calculator:
1) To find 1%, divide by 100 (70 / 100 = 0.7) 

2) Then, multiply by the % you need (0.7 x 13 = 9.1) 

Therefore 70 - 9.1 = 60.9 

If you don't have a calculator: 
1) Find 50%, 10%, 5%, and 1%. With these, you can make any % you need. 
50% = X / 2 (= 35) 
10% = X / 10 (= 7) 
5% = first, do 10%, then divide that by 2 (= 3.5) 
1% = X / 100 (= 0.7) 
Where X is the number you want to work out % for 

2) Add those % up to get the amount you want 
10% + 1% + 1% +1% = 13% 
7 + 0.7 + 0.7 + 0.7 = 9.1 

Therefore 70 - 9.1 = 60.9
Hope that helps :)

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