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Makovka662 [10]

3 years ago

6

Which recursive definition could be used to generate the sequence {3, 6, 3, 6, 3,...}? A) a1 = 3 and an = an-1 + 3n

1 answer:

barxatty [35]3 years ago

8 0

The answer is choice C

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You start off with the first term being a1 = 3

Then you add on 3 to get to 6 as the second term

The next thing to do is add on -3, which is the same as subtracting 3, to get back to 3 once more.

This pattern continues on forever. To ensure you add on either +3 or -3 to the previous term, we add on the general term 3*(-1)^n

Overall, the nth term is found by a(n) = a(n-1) + 3*(-1)^n, which is what choice C shows. I'm assuming your teacher meant to write 3(-1)^n instead of 3(-1)n

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Bogdan [553]

Given:

In the given circle O, BC is diameter, OA is radius, DC is a chord parallel to chord BA and $m\angle BCD=30^\circ$ .

To find:

The $m\angle AOB$ .

Solution:

If a transversal line intersect two parallel lines, then the alternate interior angles are congruent.

We have, DC is parallel to BA and BC is the transversal line.

$\angle OBA\cong \angle BCD$ [Alternate interior angles]

$m\angle OBA=m\angle BCD$

$m\angle OBA=30^\circ$

In triangle AOB, OA and OB are radii of the circle O. It means OA=OB and triangle AOB is an isosceles triangle.

We know that base angles of an isosceles triangle are congruent.

$\angle OAB\cong \angle OBA$ [Base angles of an isosceles triangle]

$m\angle OAB=m\angle OBA$

$m\angle OAB=30^\circ$

In triangle AOB,

$m\angle OAB+m\angle OBA+m\angle AOB=180^\circ$

$30^\circ+30^\circ+m\angle AOB=180^\circ$

$60^\circ+m\angle AOB=180^\circ$

$m\angle AOB=180^\circ-60^\circ$

$m\angle AOB=120^\circ$

Therefore, the measure of angle AOB is 120 degrees.

6 0

3 years ago

A garden snail moves 1/6 foot in 1/3 hours. what is the unit rate?

fiasKO [112]

1 hour=1/6+1/6+1/6
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3 years ago

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Trava [24]

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7 0

3 years ago

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ziro4ka [17]

Answer:

The correct answer would be $a=\sqrt{c^2-b^2}$

Step-by-step explanation:

given $a^2+b^2=c^2$ we want to solve for a

How?

We can do this by using basic algebra ( isolating the variable (a ))

Step 1 subtract $b^2$ from each side

$a^2+b^2-b^2=a^2\\c^2-b^2=c^2-b^2$

now we have $a^2=c^2-b^2$

step 2 take the square root of each side

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we're left with $a=\sqrt{c^2-b^2}$

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3 years ago

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vazorg [7]

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