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melomori [17]
3 years ago
6

WILL MARK CORRECT ANSWER BRAINLIEST

Mathematics
1 answer:
Andrew [12]3 years ago
4 0
F(x) = 3x I think that will be the right answer.
You might be interested in
If n and k are positive integers, is n divisible by 6 ? (1) n = k(k + 1)(k – 1) (2) k – 1 is a multiple of 3.
xz_007 [3.2K]

Answer: yes, it is

Step-by-step explanation:

A number is divisible by 6 if it is divisible by 2 and 3 simultaniously.

n = k(k+1)(k-1)

If k-1 is a multiple of 3, n is divisible by 3, so one of the requirements is ok.

Now, if k-1 is a multiple of 3, it can be even or odd.

if k-1 is even, then it is divisible by 2 and as it is divisible by 3 as well, n is divisible by 6

if k-1 is odd, then k and k+1 is even, hence, divisible by 2.

As n = k(k+1)(k-1), n is also divisible by 6.

4 0
4 years ago
X 3,4,5,6
aliina [53]

Answer:

Linear

Step-by-step explanation:

We are adding by 5 everytime so it is linear!

9 + 5 = 14

14 + 5 = 19

19 + 5 = 24

4 0
3 years ago
Write sin 81 degrees in terms of cosine
NARA [144]

Answer:

Sine, cosine and tangent formulas are used in RIGHT triangles. So you have one 90 degree angle and two acute angles that add up to 90 degrees (they are complimentary). So the cosine of one angle would be the same as the sine of the compliment to that angle... and vice versa. To answer this question, sin 81 = cos 9, since 81 and 9 are the two acute angles of the triangle and they both add up to 90 degrees. "Cos 9" is the answer when you wish to write "sin 81" in terms of "cosine."

Please give me brainiest

8 0
3 years ago
What is the polar form of -3+ v3i?
Softa [21]
  • z=-3+√3i
  • a=-3,b=√3

So

\\ \rm\Rrightarrow r=\sqrt{a^2+b^2}

\\ \rm\Rrightarrow r=\sqrt{(-3)^2+(\sqrt{3})^2}

\\ \rm\Rrightarrow r=\sqrt{9+3}

\\ \rm\Rrightarrow r=\sqrt{12}

\\ \rm\Rrightarrow r=2\sqrt{3}

Find theta

\\ \rm\Rrightarrow \theta=tan^{-1}(\dfrac{y}{x})

\\ \rm\Rrightarrow \theta=tan^{-1}(\dfrac{\sqrt{3}}{3})

\\ \rm\Rrightarrow \theta=tan^{-1}(\dfrac{-1}{\sqrt{3}})

\\ \rm\Rrightarrow \theta=\dfrac{5\pi}{6}

Polar form

\\ \rm\Rrightarrow r(cos\theta+isin\theta)

\\ \rm\Rrightarrow 2\sqrt{3}(cos\dfrac{5\pi}{6}+isin\dfrac{5\pi}{6})

7 0
2 years ago
Please help me please i really need help please
ziro4ka [17]
B is the answer i I just took it
6 0
3 years ago
Read 2 more answers
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