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3 years ago

Write 61 using Egyptian and Babylonian numbers.

Mathematics

1 answer:

Marrrta [24]3 years ago

4 0

Answer:

61 in Egyptian numeral is ∩∩∩∩∩∩l

61 is written as - т т

Step-by-step explanation:

Egyptian numeral:

∩ mean 10

l = 1

for 60 we can use 6 number of ∩∩∩∩∩

i.e.

therefore 61 in Egyptian numeral is ∩∩∩∩∩∩l

Babylonian numeral : Basically Babylonian number system is 60 based instead of 10.

there are total number of 59 numerals made up by two symbol only.

61 is written as - т т with a space between two symbol

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Find the value of the 60th term of the following sequence: 33, 28, 23, 18

Otrada [13]

Step-by-step explanation:

The first term of the AP i.e a = 33

and common difference i.e d = 28-33 = -5

So, 60th term = a +(n-1)d = a+ 59d

= 33 + 59( -5)

= 33- 295

= - 262

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2 years ago

A 3-foot wife wood deck will be built around rectangular shaped pool that measures 15feet by 30feet. The resulting shape is also

lisov135 [29]

Answer:

$340

Step-by-step explanation:

The pool is 5 yards wide by 10 yards long, so adding the deck will make the total area 7 yards wide by 12 yards long, or ...

(7 yd)·(12 yd) = 84 yd^2

The pool itself is ...

(5 yd)·(10 yd) = 50 yd^2

so the area of the deck is ...

84 yd^2 -50 yd^2 = 34 yd^2

Wood for each of those square yards costs $10, so the total cost is ...

($10/yd^2)·(34 yd^2) = $340

6 0

3 years ago

Plz help I'll appreciate it

Nutka1998 [239]

The unit rate is 6 I think

5 0

2 years ago

Jonah calculates that the perimeter of the rectangle below is 6 1/4x - 3. Explain his error and include the correct expression.

Natalka [10]

Answer:

Jonah left out the other two sides!

3 0

2 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago