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aleksandr82 [10.1K]
3 years ago
13

Adele opens an account with $140 and deposits $45 a month. Kent opens an account with $50 and also deposits $45 a month. Will th

ey have the same amount in their accounts at any point? If so, in how many months and how much will be in each account?
Mathematics
1 answer:
EastWind [94]3 years ago
7 0

Answer:

They will never have the same amount of money

Step-by-step explanation:

It goes in at the same rate and the same amount so, one person will always be $90 ahead of the other.

I also had this problem.

Your Welcome!

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Convert 5x + 3y = 11 to slope-intercept form.
ser-zykov [4K]

Answer:

Slope = -3.333/2.000 = -1.667

 x-intercept = 11/5 = 2.20000

 y-intercept = 11/3 = 3.66667

Steps below

Step-by-step explanation:

Step  1  :

Equation of a Straight Line

Calculate the Y-Intercept

Calculate the X-Intercept

Calculate the Slope

Done!

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4 years ago
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For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva
sergij07 [2.7K]

Given

we are given a function

f(x)=x^2+5

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

\Delta x=\frac{b-a}{n}

and the endpoints are given by

a+k.\Delta x,\text{ for }0\leq k\leq n

For k=0 and k=n, we get

\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}

Each rectangle has width and height as

\Delta x\text{ and }f(x_k)\text{ respectively.}

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)

Here

f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}\Delta x=\frac{5-0}{n}=\frac{5}{n}x_k=0+k.\Delta x=\frac{5k}{n}f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5

Now Area=

\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}

So the required area is 66.6 sq units.

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tamaranim1 [39]

Answer:

Step-by-step explanation:

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Range: [2,6]

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