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ivanzaharov [21]
3 years ago
10

Help i will give brainliest

Mathematics
1 answer:
shusha [124]3 years ago
3 0

Answer:

The ratio between A and B is 30/1.

Step-by-step explanation:

In math, ratios are the relationships between two quantities using division. They want us to find the ratio of A (3 * 10^-6) to B (0.0000001). While the quantity of A is in scientific notation, the quanitity of B is in decimal notation.

Scientific notation is a way to notate very large or very small numbers in the form of a * 10^m, where 1 ≤ a < 10.

Decimal notation is the regular notation we use for rational numbers that are not fractions or percentages.

I will do the math in scientific notation. First, I will convert B from decimal notation to scientific notation. To do this, I must first find the decimal point in B.

0.0000001

Then, I move the decimal place right until the a number is between 1 and 10.

1

But we can't just make such a small number into 1 like nothing! I have to notify the change in a using powers of 10. SInce I moved the decimal place 7 places and the number got smaller, I will say that 1 can be multiplied by 10^-7

1 * 10^-7.

Now I can find the ratio between A and B. First, I write the ratio as is.

(3 * 10^-6)/(1 * 10^-7)

Next, I can rearrange the a number and the powers of 10 with each other.

= (3/1)(10^-6/10^-7)

Simplify.

= (3)(10)

= 30

The ratio between A and B is 30/1.

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100 points , please help. I am not sure if I did this correct if anyone can double-check me thanks!
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Step-by-step explanation:

\lim_{n \to \infty} \sum\limits_{k=1}^{n}f(x_{k}) \Delta x = \int\limits^a_b {f(x)} \, dx \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times k

In this case we have:

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b − a = 3

a = 1

b = 4

So the integral is:

∫₁⁴ √x dx

To evaluate the integral, we write the radical as an exponent.

∫₁⁴ x^½ dx

= ⅔ x^³/₂ + C |₁⁴

= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)

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If ∫₁⁴ f(x) dx = e⁴ − e, then:

∫₁⁴ (2f(x) − 1) dx

= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx

= 2 (e⁴ − e) − (x + C) |₁⁴

= 2e⁴ − 2e − 3

∫ sec²(x/k) dx

k ∫ 1/k sec²(x/k) dx

k tan(x/k) + C

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k tan(π/(2k)) + C − (k tan(0) + C)

k tan(π/(2k))

Setting this equal to k:

k tan(π/(2k)) = k

tan(π/(2k)) = 1

π/(2k) = π/4

1/(2k) = 1/4

2k = 4

k = 2

8 0
4 years ago
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