The scale on the map is 1:250000

1 answer:

6 0

Scale 1: 250,000 literally means for 1cm you calculated on the map it is magnified by 250,000 times. So if that road is 4-6cm, you just multiply it by 250,000 times and you get 1,150,000 cm.
Since you wanted to be in Km. the 100 cm is 1 meter and 1000 meters is 1 km. Therefore, 1,150,000/ 100 x 1000 = 11.5 km . Answer: 11.5 km.

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The sequence 2, 3, 5, 6, 7, 10, 11, $\ldots$ contains all the positive integers from least to greatest that are neither squares

mariarad [96]

Answer:

1041

Step-by-step explanation:

Among the numbers 1–1000, there are 31 squares, so the sequence will extend to at least 1031. In those added numbers, there is another square (1024), so the sequence must extend to at least 1032.

There are 7 more numbers that are cubes, but not squares, so the sequence must extend to at least 1039.

And there are 2 additional numbers that are 5th powers that are not squares or cubes. Compensating for the removal of these numbers extends the end of the sequence to 1041.

There are no numbers in this range that are both cubes and 5th powers and that have not already been accounted for. (The only 15th power is 1.)

Hence, the 1000th number in the sequence is 1041.

_____

This result is verified by a computer program that listed the numbers.

3 0

4 years ago

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$