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

5 0

3 years ago

Is 4+(3×y) (4+3)×y equivalent or 10-(6÷y) (10-6)÷y

Daniel [21]

They don't seem equivalent, the first set yields 4+3xy and 7xy. The second set yields 10-(6/y) and 4/y. R u sure u typed in the question correctly?

7 0

3 years ago

Write the fraction as a sum of unit fractions

OLEGan [10]

Answer:

1/10+1/10+1/10

1/6+1/6+1/6+1/6+1/6+1/6 or simply 1

Step-by-step explanation:

7 0

4 years ago

F(x) = (x + 2) (x - 7)3

mote1985 [20]

Answer:

y=3x^2 -15x-42

Step-by-step explanation:

8 0

3 years ago