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klemol [59]
3 years ago
8

Use the explicit formula an=a1+(n-1)

Mathematics
2 answers:
Ronch [10]3 years ago
6 0
The anwser is a hope it helps
azamat3 years ago
3 0
T_n = a_1 + (n - 1) d

<em>The sequence is:</em>
24, 31, 38,45, 52

<em>We can see from the sequence:</em>
a = 24, d = 7

<em>Find nth term:</em>
T_n = 24 + 7(n - 1)
T_n = 24 + 7n - 7
T_n = 7n + 17

<em>When  = 500,</em>
T_{500} = 7(500) + 17
T_{500} = 3517

Answer: 500th term is 3517

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Diego solved an equation by multiplying both sides of the equation by 6. Then he checked that 6 is the correct solution by subst
Nikolay [14]
Based on the given condition Diego utilized, it can be said that he used multiplication property when he multiplied both sides of the equation by 6 and then substitution property when he substituted 6 to the original equation.
(c) Multiplication property of equality and substitution property of equality
7 0
3 years ago
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
aliya0001 [1]

Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

3 0
2 years ago
Robin purchased a $160,000 182-day T-bill discounted to yield 1.96%. When he sold it 45 days later, yields had dropped to 1.82%.
Sav [38]

The amount earned by Robin is $224

  • Given the cost of 182-day T-bill = $160,000

If Robin is discounted to yield 1.96%, the amount yield is expressed as:

The yield of 1.96% = 1.96% of 160000

The yield of 1.96% = 0.0196 * 160000

The yield of 1.96% = $3136

Similarly, if the yield is dropped to 1.82%

The yield of 1.82% = 1.82% of 160000

The yield of 1.82% = 0.0182 * 160000

The yield of 1.82% = $2,912

Amount earned by Robin = $3136 - $2192

Amount earned by Robin  = $224

Hence the amount earned by Robin is $224

Learn more on discounts here: brainly.com/question/17745353

6 0
2 years ago
Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)
Svet_ta [14]
<h2>Answer:</h2>

\boxed{x=1, \y=-1, \ z=2}

<h2>Step-by-step explanation:</h2>

We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:

Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}

Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}

Step 4: solve for z, then for y, then for x:

\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}

-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}

By substituting y=-1 \ and \ z=2 into the first equation, we get the x. So:

x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}

6 0
3 years ago
Read 2 more answers
What is a 8096 rounded to 1 significant figure<br>​
guajiro [1.7K]

Answer:

Y Me Angry U Hungry Me And Me Angry Way Back together Me Y He Angry Me He Angry Me Angry:(

5 0
3 years ago
Read 2 more answers
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