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kykrilka [37]
3 years ago
9

(Q2) Determine the eccentricity, the type of conic, and the directrix for r = 9/1+cos theta

Mathematics
1 answer:
g100num [7]3 years ago
3 0

Answer:

The answer is (d) ⇒ eccentricity = 6 , hyperbola , directrix  x = 3/2

Step-by-step explanation:

∵ r = ek/(1 ± ecosФ)

∵ r = 9/(1 + 6cosФ)

∴ e = 6 ⇒ eccentricity = 6

∵ e > 1

∴ The conic is hyperbola

∵ ek = 9

∴ k = 9/6 = 3/2

∴ The directrix  x = 3/2

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Select all the expressions that are equivalent to (2)^n+³
eimsori [14]

Answer:

The expressions which equivalent to  (2)^{n+3} are:

4(2)^{n+1}  ⇒ B

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Step-by-step explanation:

Let us revise some rules of exponent

  • a^{m} × a^{m}  = a^{m+n}
  • (a^{m})^{n} = a^{m*n}

Now let us find the equivalent expressions of  (2)^{n+3}

A.

∵ 4 = 2 × 2

∴ 4 =  2^{2}

∴  (4)^{n+2} =  (2^{2})^{n+2}

- By using the second rule above multiply 2 and (n + 2)

∵ 2(n + 2) = 2n + 4

∴  (4)^{n+2} =  (2)^{2n+4}  

B.

∵ 4 = 2 × 2

∴ 4 =  2²

∴  4(2)^{n+1} = 2² ×  (2)^{n+1}

- By using the first rule rule add the exponents of 2

∵ 2 + n + 1 = n + 3

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C.

∵ 8 = 2 × 2 × 2

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∴  8(2)^{n} = 2³ ×  (2)^{n}

- By using the first rule rule add the exponents of 2

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D.

∵ 16 = 2 × 2 × 2 × 2

∴ 16 = 2^{4}

∴  16(2)^{n} = 2^{4}  ×  (2)^{n}

- By using the first rule rule add the exponents of 2

∵ 4 + n = n + 4

∴  16(2)^{n} =  (2)^{n+4}

E.

(2)^{2n+3} is in its simplest form

The expressions which equivalent to  (2)^{n+3} are:

4(2)^{n+1}  ⇒ B

8(2)^{n} ⇒ C

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