7. Adrienne and her brother John each inherit $7,000 that they use to open savings accounts with Adrienne's account earns 4.7% annual compound interest and John's account earns 4.7% annual
simple interest. Neither makes any additional deposits or withdrawals for 10 years.
a. Who's account will be worth more at the end of 10 years?
b. How much more will the account be worth?

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Ответ:

rayraye

01.09.2022

Mathematics

This question is, in other words, asking when the graph of $y = x^3 - 7x^2 + 36$ is negative (below the x-axis).we have to factor the leftside first to find the zeros. the zeros of the function $y = x^3 - 7x^2 + 36$ will be where the graph will cross the x-axis.rational root theorem tells us that possible factors of this polynomial are possible factors of the constant term.factors of 36: $\pm 1, \pm 2, \pm 3, \pm4, /tex]we use rational root theorem with remainder theorem, which tells us that putting a value [tex]a$ in a function gets us the remainder of dividing that function by $x-a$ test those factors to find a value that makes the function zero so that we can find the right binomial to divide by to get a remainder of zero. $y(1) = (1)^3 - 7(1)^2 + 36 = 30$ : no $y(-1) = (-1)^3 - 7(-1)^2 + 36 = 28$ : no $y(2) = (2)^3 - 7(2)^2 + 36 = 16$ : no $y(-2) = (-2)^3 - 7(-2)^2 + 36 = 0$ : yessince putting in $x=-2$ makes the function zero, then dividing $x^3 - 7x^2 + 36$ by $x+2$ will have remainder of zero; this tells us that $x+2$ is factor.division is performed as attachment, with both synthetic and long division as options. either way, quotient of $x^2-9x+18$ . the division statement for this is $\dfrac{x^3 - 7x^2 + 36}{x+2} = x^2-9x+18$ multiplying both sides by $x+2$ : $x^3 - 7x^2 + 36 = (x+2)(x^2-9x+18)$ therefore $\begin{aligned} x^3 - 7x^2 + 36 & \ \textless \ 0 \\ (x+2)(x^2-9x+18) & \ \textless \ 0 \end{aligned}$ $x^2-9x+18$ is factorable using normal methods. two numbers that multiply to get +18 and add to get -9: -6 and -3 $\begin{aligned}(x+2)(x^2-9x+18)\ \textless \ 0 \\ (x+2)(x-3)(x-6)\ \textless \ 0 \end{aligned}$ you can either graph this polynomial or use region testing.we have the following regions to test: region 1: $x \ \textless \ -2$ region 2: $-2 \ \textless \ x \ \textless \ 3$ region 3: $3\ \textless \ x\ \textless \ 6$ region 4: $x \ \textgreater \ 6$ we use the zeros as region boundaries because the zeros are the only place a polynomial function can cross the x-axis, which determines if the result is positive or negative. testing a value in a region formed by the zeros determines if the result is positive or negative for that entire region.pick a value for each region to test if the results is negative or positive. if it is negative, then the region does satisfy $x^3 - 7x^2 + 36\ \textless \ 0$ .region 1: possible value to test in $x \ \textless \ -2$ is -3: $(-3)^3 - 7(-3)^2 + 36 = -54$ so $x\ \textless \ -2$ is a solution.region 2: possible value to test in $-2 \ \textless \ x \ \textless \ 3$ is 0 $(0)^3 - 7(0)^2 + 36 = 36$ this result is positive, so this region is not a solution.region 3: possible value to test in $3\ \textless \ x\ \textless \ 6$ is 4 $(4)^3 - 7(4)^2 + 36 = -12$ so $3\ \textless \ x\ \textless \ 6$ is a solution.region 4: possible value to test in $x \ \textgreater \ 6$ is 7. $(7)^3 - 7(7)^2 + 36 = 36$ this result is positive, so this region is not a solution.so the values that satify are: $x \ \textless \ -2,\ 3 \ \textless \ x \ \textless \ 6$

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