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kherson [118]
3 years ago
9

Julio wants to calculate 200,000 300,000. When he used

Mathematics
2 answers:
IRISSAK [1]3 years ago
4 0
Answer: D

Just add 10 zeros behind the 6.
You get 60,000,000,000

Checking this with a calculator verifies this is the correct answer.
statuscvo [17]3 years ago
3 0

Answer:

D = 60,000,000,000

Step-by-step explanation:

When "E" appears on a calculator , it means exponential and it is used to denote numbers that are too big or small to appear on the calculator screen in their decimal form.

Therefore 6E-10 means 6 exponential 10 , mathematically written as  6^10

which is 60,000,000,000

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Consider the following theorem. Theorem If f is integrable on [a, b], then b a f(x) dx = lim n→[infinity] n i = 1 f(xi)Δx where
mel-nik [20]

Split up the interval [1, 9] into <em>n</em> subintervals of equal length (9 - 1)/<em>n</em> = 8/<em>n</em> :

[1, 1 + 8/<em>n</em>], [1 + 8/<em>n</em>, 1 + 16/<em>n</em>], [1 + 16/<em>n</em>, 1 + 24/<em>n</em>], …, [1 + 8 (<em>n</em> - 1)/<em>n</em>, 9]

It should be clear that the left endpoint of each subinterval make up an arithmetic sequence, so that the <em>i</em>-th subinterval has left endpoint

1 + 8/<em>n</em> (<em>i</em> - 1)

Then we approximate the definite integral by the sum of the areas of <em>n</em> rectangles with length 8/<em>n</em> and height f(x_i) :

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx \approx \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right)

Take the limit as <em>n</em> approaches infinity and the approximation becomes exact. So we have

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx = \lim_{n\to\infty} \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right) \\\\ = \lim_{n\to\infty} \frac8n \sum_{i=1}^n \left(1+\frac{16}n(i-1)+\frac{64}{n^2}(i-1)^2-4-\frac{32}n(i-1)+6\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=1}^n \left(64(i-1)^2-16n(i-1)+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=0}^{n-1} \left(64i^2-16ni+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(64\sum_{i=0}^{n-1}i^2 - 16n\sum_{i=0}^{n-1}i + 3n^2\sum{i=0}^{n-1}1\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{64(2n-1)n(n-1)}{6} - \frac{16n^2(n-1)}{2} + 3n^3\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{49n^3}3-24n^2+\frac{32n}3\right) \\\\= \lim_{n\to\infty} \frac{8\left(49n^2-72n+32\right)}{3n^2} = \boxed{\frac{392}3}

3 0
3 years ago
What is the area, in square millimeters, of a right triangle with a hypotenuse of 68 millimeters and a leg of 60 millimeters?
Paladinen [302]
First, we have to solve the sides of the right triangle by using the Pythagorean theorem: a² + b² = c² where c is the length of the hypotenuse. Then, we can use the legs as length and height of the triangle and find the area by 1/2(base x height).

x² + 60² = 68²
x² + 3600 = 4624
x² = 1024
x = √1024
x = 32 = length of the other side

A = (1/2)(32) (60) = 960 mm²
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10:21
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Answer:

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