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serg [7]
3 years ago
8

1.64*10^0/2.0*10^2 in scientific notation

Mathematics
1 answer:
jarptica [38.1K]3 years ago
5 0
<span>1.64*10^0/2.0*10^2 in scientific notation

</span>1.64 * 10^0 : 2.0 * 10^2= (1.64:2.0) * 10^0:10^2= 0.82 * 10 ^{0-2} =  \\  \\ 0.82 * 10 ^{-2} =0.82 * \frac{1}{10^2} =0.82 * \frac{1}{100} =0.0082=  \boxed{ 8.2 * 10^{-3} }<span>
</span>
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Find the nature of the root
GREYUIT [131]

Answer:

1) Real and same.

2) Real and distinct

3) Real and distinct

4) Real and distinct

5) Real and distinct

6) Real and distinct

7) Real and distinct

8) Real and distinct

Step-by-step explanation:

If a quadratic equation is in the form of ax² + bx + c = 0, then Discriminant of the equation D = b² - 4ac, which governs the nature of roots the equation has.

If D > 0, then there will be two different real roots.

If D = 0, then two same and real roots.

If D < 0, then two distinct but imaginary roots.

Now, 1) x² + 6x + 9 = 0 has D = 6² - 4 × 9 × 1 = 0. So, there will be two same and real roots.

2) 5x² - x = 4x² + 2

⇒ x² - x - 2 = 0

It has D = (-1)² - 4 × 1 × (-2) = 9. Therefore, the roots will be two distinct and real.

3) \frac{2}{x} + \frac{3}{x} = x - 4

⇒ 5 = x² - 4x

⇒ x² - 4x - 5 = 0.

So, D = (-4)² - 4 × (1) × (- 5) = 36. So, the equation has two real and distinct rools.

4) x(x - 5) = 4(5 - x)

⇒ x² - x - 20 = 0

Hence, D = (-1)² - 4 × 1 × (-20) = 81

So, the roots will be real and distinct.

5) x² + 7x + 1 = 0 has D = 7² - 4 × 1 × 1 = 45

So, the roots will be real and distinct.

6) 2x² + 9x + 3 = 0, has D = 9² - 4 × 2 × 3 = 57.

Hence, the roots will be real and distinct.

7) 5x² - 6 = 13x

⇒ 5x² - 13x - 6 = 0

So, D = (-13)² - 4 × 5 × (-6) = 289

So, the roots will be real and distinct.

8) x² - x = 3(x + 7)

⇒ x² - 4x - 21 = 0

It has D = (-4)² - 4 × 1 × (-21) = 100

So, the roots will be real and distinct. (Answer)

6 0
3 years ago
Can two obtuse angles be complementary
Norma-Jean [14]

Complementary angles are angles that add up to 90 degrees. 

Obtuse angles are greater than 90 degrees, so there's no way
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either one already is. 

The answer to the question is 'No'.
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At Arnold's Arcade, a child can buy 21 tokens for $6. At this rate, how many tokens would $20 get someone?
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Based on the information given, the number of tokens that will be bought if a child has $20 will be 70.

Since a child can buy 21 tokens for $6, the cost of one token will be:

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Read related link on:

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Math homework due tomorrow
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