The quadratic $x^2+\frac{3}{2}x-1$ has the following unexpected property: the roots, which are $\frac{1}{2}$ and $-2$, are one $\textit{less}$ than the final two coefficients. Now find a quadratic with leading term $x^2$ such that the final two coefficients are both non-zero, and the roots are one $\textit{more}$ than these coefficients.

i believe the correct answer is option 4.principalities were small territories that were run by russian rulers.

which statement correctly defines principalities in northern asia in the 1200s? principalities were small territories that were run by russian rulers.

hope i could ! : )

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places

answer: (3,11) is your answer.

i think its doubled. let me know

step-by-step explanation: