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3 years ago

How do you spell 4 billion in scientific notation

Mathematics

1 answer:

Tema [17]3 years ago

3 0

Answer:

<h2>4 bilion = 4 × 10⁹</h2>

Step-by-step explanation:

The scientific notation:

$a\cdot10^k$

where

$1\leq a$

4 billion:

in USA: 4,000,000,000

in Europe: 4 000 000 000 000

$4\underbrace{000000000}_9=4\cdot10^9$

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How do you write 8.216 in scientific notation?

Igoryamba

How you write 8.216 in scientific notation; 8.216 * 10 ^ 3

10 ^ 3 = 10 to the 3rd power.

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3 years ago

An article which cost money 150 was sold at a loss of 5%.what was the selling price?

Hunter-Best [27]

Answer:

142.5

Step-by-step explanation:

150/x=100/95

95*150/100=

142.5

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3 years ago

Which of the following hace a value equal to |37|? Select all that apply

daser333 [38]

B, D, and E
The absolute value of -37 is 37.
The absolute value of 37 is also 37.
When you distribute the negative in the parenthesis the -37 becomes positive.

3 0

3 years ago

What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c

zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

4 0

3 years ago

Find center and radius<br> (x-3)^2+y^2=8

Flauer [41]

Answer:

<h2>Center: (3, 0)</h2><h2>Radius: 2√2</h2>

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the equation:

$(x-3)^2+y^2=8\to(x-3)^2+(y-0)^2=(\sqrt8)^2$

Therefore

the center: $(3,\ 0)$

the radius: $r=\sqrt8\to r=\sqrt{4\cdot2}\to r=\sqrt4\cdot\sqrt2\to r=2\sqrt2$

4 0

3 years ago