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3 years ago

15

Estimate. Then find the difference. 253,495-48,617=

2 answers:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

Difference of the two numbers will be 204878.

Step-by-step explanation:

The given expression is (253495 - 48617)

We have to calculate the difference of two numbers of the expression.

We can rewrite the numbers as

253495 = 2×100000 + 53×10000 + 4×100 + 95

048617 = 0×100000 + 48×10000 + 6×100 + 17

We can subtract these numbers to find difference

253495 - 048617 = (2×100000 - 0×100000) + (53×1000 - 48×1000) + (4×100 - 6×100) + (95 - 17)

= 100000(2 - 0) + 1000(53 - 48) + 100(4 - 6) + 78

= 2×100000 + 5×1000 - 2×100 + 78

= 200000 + 5000 - 200 + 78

= 2050000 - 122

= 204878

BartSMP [9]3 years ago

5 0

300000-50000=250000

...................

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2 years ago

Match each series with the equivalent series written in sigma notation

PIT_PIT [208]

Answer:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

$3 + 12 + 48 + 192 + 768$

The expression can be written as:

$3 \to 3*4^{0$ --- 0

$12 \to 3 * 4^{1$ ---- 1

$48 \to 3 * 4^{2$ --- 2

$192 \to 3 * 4^{3$ ---- 3

$768 \to 3 * 4^{4$ ---- 4

For the nth term, the expression is:

$Term = 3 * 4^{n$ ---- n

So, the summation is:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

Solving (b):

$4 + 32 + 256 + 2048 + 16384$

The expression can be written as:

$4 \to 4 * 8^0$ --- 0

$32 \to 4 * 8^1$ ---- 1

$256 \to 4 * 8^2$ --- 2

$2048 \to 4 * 8^3$ ---- 3

$16384 \to 4 * 8^4$ ---- 4

For the nth term, the expression is:

$Term \to 4 * 8^n$ ---- n

So, the summation is:

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

Solving (c):

$2 + 6 + 18 + 54 + 162$

The expression can be written as:

$2 \to 2 * 3^0$ --- 0

$6 \to 2 * 3^1$ ---- 1

$18 \to 2 * 3^2$ --- 2

$54 \to 2 * 3^3$ ---- 3

$162 \to 2 * 3^4$ ---- 4

For the nth term, the expression is:

$Term \to 2 * 3^n$ ---- n

So, the summation is:

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

Solving (d):

$3 + 15 + 75 + 375 + 1875$

The expression can be written as:

$3 \to 3 * 5^0$ --- 0

$15 \to 3 * 5^1$ ---- 1

$75 \to 3 * 5^2$ --- 2

$375 \to 3 * 5^3$ ---- 3

$1875 \to 3 * 5^4$ ---- 4

For the nth term, the expression is:

$Term \to 3 * 5^n$ ---- n

So, the summation is:

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

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