$Suppose~that~the~number~is~x.\\According~to~question,\\x^2-140=3x\\or, x^2-3x-140=0\\From~the~quadratic~formula,\\x = \frac{-(-3)+\sqrt{(-3)^2-4(1)(-140)} }{2(1)} ~and ~x= \frac{-(-3)-\sqrt{(-3)^2-4(1)(-140)} }{2(1)} \\or, x = \frac{3+\sqrt{9+560} }{2}~and~x=\frac{3-\sqrt{9+569} }{2}\\or, x = \frac{3+\sqrt{569} }{2}~and~x=\frac{3-\sqrt{569} }{2}\\So,~the~required~negative~number~is~\frac{3-\sqrt{569} }{2}.$

5 0

3 years ago

A person invested $7,400 in an account growing at a rate allowing the money to

Reil [10]

Answer:

well the money doubles every 6 years but its only been 4 so

7,400 x 4 = 29,600

$29,600 is the answer

8 0

2 years ago

Solve the system using an equivalent system

DerKrebs [107]

Add y to both sides in 1st equation
-2x=y
sub -2x for y in2nd equation
3x-(-2x)=10
3x+2x=10
5x=10
divide 5
x=2

sub back
-2x=y
-2(2)=y
-4=y

(x,y)
(2,-4)
C is ans

8 0

3 years ago