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2 years ago

Question Solve. 2 1/3−1/2x=−2/3 Enter your answer in the box.

Mathematics

1 answer:

Tcecarenko [31]2 years ago

6 0

2/3 - 1/2x = -2/3 then the value of x is 6 .

The algebraic expression should often take one of the following forms: addition, subtraction, multiplication, or division. Bring the variable to the left and the remaining values to the right to determine the value of x. To determine the outcome, simplify the values.

The steps to use the calculator to determine x's value are as follows:

Fill in the multiplicand and product fields with the numbers (Integer/Decimal Numbers).
To obtain the result, press the "Solve" button now.
The output field will show the dividend or the x value.

given equation is

2/3 - 1/2x = -2/3

⇒7/3-x/2 = -2/3

⇒7/3+ 2/3 =x/2

⇒9/3 = x/2

⇒ x=3 x 2

⇒ x=6

therefore the value of the x is 6

To know more about how to find x values visit :

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$20000 is invested in an account that earned 6% p.A. Compounding yearly for 3 years. The interest rate then went up to 8% p.A. F

GuDViN [60]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}$

$\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762$

4 0

4 years ago

Can someone help me with 11 A and B?

Elan Coil [88]

For 11a its 2:10 because how you would come up with it is by placing pablos free minutes first and then putting pablos paid minutes and 11b is8:25 because once again you would place sams free minutes first and then his paid minutes

6 0

3 years ago

Determine the above sequence converges or diverges. If the sequence converges determine its limit

marshall27 [118]

Answer:

This series is convergent. The partial sums of this series converge to $\displaystyle \frac{2}{3}$ .

Step-by-step explanation:

The $n$ th partial sum of a series is the sum of its first $n\!\!$ terms. In symbols, if $a_n$ denote the $n\!$ th term of the original series, the $\! n$ th partial sum of this series would be:

$\begin{aligned} S_n &= \sum\limits_{k = 1}^{n} a_k \\ &= a_1 + a_2 + \cdots + a_{k}\end{aligned}$ .

A series is convergent if the limit of its partial sums, $\displaystyle \lim\limits_{n \to \infty} S_{n}$ , exists (should be a finite number.)

In this question, the $n$ th term of this original series is:

$\displaystyle a_{n} = \frac{{(-1)}^{n+1}}{{2}^{n}}$ .

The first thing to notice is the ${(-1)}^{n+1}$ in the expression for the $n$ th term of this series. Because of this expression, signs of consecutive terms of this series would alternate between positive and negative. This series is considered an alternating series.

One useful property of alternating series is that it would be relatively easy to find out if the series is convergent (in other words, whether $\displaystyle \lim\limits_{n \to \infty} S_{n}$ exists.)

If $\lbrace a_n \rbrace$ is an alternating series (signs of consecutive terms alternate,) it would be convergent (that is: the partial sum limit $\displaystyle \lim\limits_{n \to \infty} S_{n}$ exists) as long as $\lim\limits_{n \to \infty} |a_{n}| = 0$ .

For the alternating series in this question, indeed:

$\begin{aligned}\lim\limits_{n \to \infty} |a_n| &= \lim\limits_{n \to \infty} \left|\frac{{(-1)}^{n+1}}{{2}^{n}}\right| = \lim\limits_{n \to \infty} {\left(\frac{1}{2}\right)}^{n} =0\end{aligned}$ .

Therefore, this series is indeed convergent. However, this conclusion doesn't give the exact value of $\displaystyle \lim\limits_{n \to \infty} S_{n}$ . The exact value of that limit needs to be found in other ways.

Notice that $\lbrace a_n \rbrace$ is a geometric series with the first term is $a_0 = (-1)$ while the common ratio is $r = (- 1/ 2)$ . Apply the formula for the sum of geometric series to find an expression for $S_n$ :

$\begin{aligned}S_n &= \frac{a_0 \cdot \left(1 - r^{n}\right)}{1 - r} \\ &= \frac{\displaystyle (-1) \cdot \left(1 - {(-1 / 2)}^{n}\right)}{1 - (-1/2)} \\ &= \frac{-1 + {(-1 / 2)}^{n}}{3/2} = -\frac{2}{3} + \frac{2}{3} \cdot {\left(-\frac{1}{2}\right)}^{n}\end{aligned}$ .

Evaluate the limit $\displaystyle \lim\limits_{n \to \infty} S_{n}$ :

$\begin{aligned} \lim\limits_{n \to \infty} S_{n} &= \lim\limits_{n \to \infty} \left(-\frac{2}{3} + \frac{2}{3} \cdot {\left(-\frac{1}{2}\right)}^{n}\right) \\ &= -\frac{2}{3} + \frac{2}{3} \cdot \underbrace{\lim\limits_{n \to \infty} \left[{\left(-\frac{1}{2}\right)}^{n} \right] }_{0}= -\frac{2}{3}\end{aligned}}_$ .

Therefore, the partial sum of this series converges to $\displaystyle \left(- \frac{2}{3}\right)$ .

8 0

3 years ago

Solve the following system of equations.

Lerok [7]

Answer:

x=5 and y=-2

Step-by-step explanation:

3x+5y=5

-(7x+5y=25)

-4x=-20

x=5

plug 5 into one of the equations. 3(5)+5y=5

y= -2

3 0

3 years ago

(fg)-1(5)=g-1f-1(5)

larisa86 [58]

Answer:

f = g/ g+1

Step-by-step explanation:

Remove Parenthesis.

fg - 1 x 5 = g - 1 x f - 1 x 5

Cancel -1 x 5 on both sides.

fg = g - 1 x f

Simplify 1 x f to f.

fg = g - f

Add f to both sides.

fg + f = g

Factor out the common term f.

f (g + 1) = g

Divide both sides by g + 1

f = g/ g + 1

6 0

4 years ago