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NeX [460]
3 years ago
12

If I have 500 dollars how much I need to make 1800 dollars

Mathematics
2 answers:
insens350 [35]3 years ago
8 0

You have $500 and want/need $1800. You would subtract the amount you want with the amount you have.

1800-500=1300

You need to acquire $1300

iVinArrow [24]3 years ago
5 0
Youd need to make a total of 13,000
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1!+2(2!)+3(3!)+...+n(n!)=(n+1)!-1<br><br><br><br><br>​
anyanavicka [17]

Step-by-step explanation:

OK, let's assume it this way:

<em>Sn=1.1!+2.2!+3.3!+...+n.n!</em><em>=</em><em>(</em><em>2</em><em>‐</em><em>1</em><em>)</em><em>.</em><em>1</em><em>!</em><em>+</em><em>(</em><em>3</em><em>-</em><em>1</em><em>)</em><em>.</em><em>2</em><em>!</em><em>+</em><em>(</em><em>4</em><em>-</em><em>1</em><em>)</em><em>3</em><em>!</em><em>+</em><em>.</em><em>.</em><em>.</em><em>+</em><em>(</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>-</em><em>1</em><em>)</em><em>.</em><em>n</em><em>!</em>

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!<em>=</em><em>(</em><em>2</em><em>.</em><em>1</em><em>!</em><em>-</em><em>1</em><em>!</em><em>)</em><em>+</em><em>(</em><em>3</em><em>.</em><em>2</em><em>!</em><em>-</em><em>2</em><em>!</em><em>)</em><em>+</em><em>(</em><em>4</em><em>.</em><em>3</em><em>!</em><em>-</em><em>3</em><em>!</em><em>)</em><em>+</em><em>.</em><em>.</em><em>.</em><em>+</em><em>(</em><em>(</em><em>n-1</em><em>)</em><em>n</em><em>!</em><em>-</em><em>n</em><em>!</em><em>)</em><em>=</em><em>(</em><em>2</em><em>!</em><em>-</em><em>1</em><em>!</em><em>)</em><em>+</em><em>(</em><em>3</em><em>!</em><em>-</em><em>2</em><em>!</em><em>)</em><em>+</em><em>(</em><em>4</em><em>!</em><em>-</em><em>3</em><em>!</em><em>)</em><em>+</em>

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!=(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+...+((n-1)n!-n!)=(2!-1!)+(3!-2!)+(4!-3!)+<em>.</em><em>.</em><em>.</em><em>+</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>!</em><em>-</em><em>n</em><em>!</em><em>=</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>!</em><em>-</em><em>1</em><em>!</em><em>=</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>!</em><em>-</em><em>1</em>

and boom problem solved

7 0
1 year ago
Solve the system of equations.
guajiro [1.7K]

Answer:

<u>Option b. (x = 3, y = 20, z = -14)</u>

Step-by-step explanation:

Given:

2x + 2y + 3z = 4

5x + 3y + 5z = 5

3x + 4y + 6z = 5

Solve using Cramer’s rule

∴ \left[\begin{array}{ccc}2&2&3\\5&3&5\\3&4&6\end{array}\right] =\left[\begin{array}{ccc}4\\5\\5\end{array}\right]

∴A = \left[\begin{array}{ccc}2&2&3\\5&3&5\\3&4&6\end{array}\right] = -1

Ax = \left[\begin{array}{ccc}4&2&3\\5&3&5\\5&4&6\end{array}\right] = -3

Ay = \left[\begin{array}{ccc}2&4&3\\5&5&5\\3&5&6\end{array}\right] =-20\\

Az = \left[\begin{array}{ccc}2&2&4\\5&3&5\\3&4&5\end{array}\right] = 14

∴ x = Ax/A = -3/-1 = 3

   y = Ay/A = -20/-1 = 20

   z = Az/A = 14/-1 = -14

<u>So, the answer is option b. (x = 3, y = 20, z = -14)</u>

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