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Orlov [11]
3 years ago
10

Question 3 of 8

Mathematics
2 answers:
mr Goodwill [35]3 years ago
7 0
C and E are the corrects answers
Lera25 [3.4K]3 years ago
3 0
C and E are the right answers.
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The following formula is for the area, A, of the curved surface area of a cone. A = pie x r x l, where r is the radius and l is
ki77a [65]
It is Pi x r x l plus Pi x r^2
5 0
3 years ago
Use spherical coordinates. let h be a solid hemisphere of radius 1 whose density at any point is proportional to its distance fr
MA_775_DIABLO [31]
Whereas the density of the hemisphere at any point is proportional to it's density, then 

 The full answer in attachment, we use the partial integral using the <span>spherical coordinates, 

and we find that </span>the mass of h = <span>πK/2</span>

7 0
3 years ago
Over the summer, for every 14 Okra seeds Dana planted, 9 plants grew. If he planted 182 seeds how many grew into plants
Black_prince [1.1K]

Answer:

117

Step-by-step explanation:

182/14=13

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7 0
3 years ago
A shop makes 3250£ in march in april it makes 3542.50£ what if the percentage increase
babunello [35]

Answer:

1.09

Step-by-step explanation:

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7 0
3 years ago
Find the smallest positive $n$ such that \begin{align*} n &amp;\equiv 3 \pmod{4}, \\ n &amp;\equiv 2 \pmod{5}, \\ n &amp;\equiv
Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number x such that taking it mod 4, 5, and 7 leaves the desired remainders:

x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6

  • Taken mod 4, the last two terms vanish and we have

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4

so we multiply the first term by 3.

  • Taken mod 5, the first and last terms vanish and we have

x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5

so we multiply the second term by 2.

  • Taken mod 7, the first two terms vanish and we have

x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7

so we multiply the last term by 7.

Now,

x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147

By the CRT, the system of congruences has a general solution

n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}

or all integers 27+140k, k\in\mathbb Z, the least (and positive) of which is 27.

3 0
3 years ago
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