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vlada-n [284]
2 years ago
12

Ginny factored 6x2 – 31x – 30 as shown:

Mathematics
2 answers:
gayaneshka [121]2 years ago
6 0

Answer:

A

Step-by-step explanation:

AVprozaik [17]2 years ago
4 0

Answer:

  (a)  Ginny made a mistake in step 1 when she identified b = 31. It should be b=–31.

Step-by-step explanation:

The value of b is the coefficient of x. In the given quadratic expression, the coefficient of x is -31, so Ginny should have started with b = -31 in step 1.

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Read 2 more answers
If you are given a3=2 a5=16, find a100.
ra1l [238]

I suppose a_n denotes the n-th term of some sequence, and we're given the 3rd and 5th terms a_3=2 and a_5=16. On this information alone, it's impossible to determine the 100th term a_{100} because there are infinitely many sequences where 2 and 16 are the 3rd and 5th terms.

To get around that, I'll offer two plausible solutions based on different assumptions. So bear in mind that this is not a complete answer, and indeed may not even be applicable.

• Assumption 1: the sequence is arithmetic (a.k.a. linear)

In this case, consecutive terms <u>d</u>iffer by a constant d, or

a_n = a_{n-1} + d

By this relation,

a_{n-1} = a_{n-2} + d

and by substitution,

a_n = (a_{n-2} + d) + d = a_{n-2} + 2d

We can continue in this fashion to get

a_n = a_{n-3} + 3d

a_n = a_{n-4} + 4d

and so on, down to writing the n-th term in terms of the first as

a_n = a_1 + (n-1)d

Now, with the given known values, we have

a_3 = a_1 + 2d = 2

a_5 = a_1 + 4d = 16

Eliminate a_1 to solve for d :

(a_1 + 4d) - (a_1 + 2d) = 16 - 2 \implies 2d = 14 \implies d = 7

Find the first term a_1 :

a_1 + 2\times7 = 2 \implies a_1 = 2 - 14 = -12

Then the 100th term in the sequence is

a_{100} = a_1 + 99d = -12 + 99\times7 = \boxed{681}

• Assumption 2: the sequence is geometric

In this case, the <u>r</u>atio of consecutive terms is a constant r such that

a_n = r a_{n-1}

We can solve for a_n in terms of a_1 like we did in the arithmetic case.

a_{n-1} = ra_{n-2} \implies a_n = r\left(ra_{n-2}\right) = r^2 a_{n-2}

and so on down to

a_n = r^{n-1} a_1

Now,

a_3 = r^2 a_1 = 2

a_5 = r^4 a_1 = 16

Eliminate a_1 and solve for r by dividing

\dfrac{a_5}{a_3} = \dfrac{r^4a_1}{r^2a_1} = \dfrac{16}2 \implies r^2 = 8 \implies r = 2\sqrt2

Solve for a_1 :

r^2 a_1 = 8a_1 = 2 \implies a_1 = \dfrac14

Then the 100th term is

a_{100} = \dfrac{(2\sqrt2)^{99}}4 = \boxed{\dfrac{\sqrt{8^{99}}}4}

The arithmetic case seems more likely since the final answer is a simple integer, but that's just my opinion...

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