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3 years ago

Which best describes the relationship between the successive terms in the sequence shown 2.4, -4.8, 9.6, -19.2

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

8 0

Answer: C) the common ratio is -2.

Step-by-step explanation:

We know that the common difference of an Arithmetic progression is given by the difference between any two the successive terms where as the common ratio is the ratio of the successive terms.

The given sequence : 2.4, -4.8, 9.6, -19.2

We can see that the term are alternating , neither increasing nor decreasing.

So we don't need to find the common difference here.

So we find the common ratio to show the relation between successive terms :-

$r=\dfrac{-4.8}{2.4}=\dfrac{9.6}{-4.8}=-2$

Hence, the common ratio is -2.

IRINA_888 [86]3 years ago

6 0

C. The common ratio is -2. Each one divided by the one before it is -2. -4.8/2=2, 9.6/-4.8=-2 and so on down the line.

Hope that helps.

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How to solve 4.3d+7.5=5.8d and get the answer

vodka [1.7K]

4.3d+7.5=5.8d
-4.3d -4.3d
7.5=1.5d
/1.5 /1.5
5=d

4 0

3 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

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3 years ago

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lora16 [44]

Answer:

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Step-by-step explanation:

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3 years ago

at a local animal shelter there are 12 small size dogs and 5 medium sized dogs everyday the small sized dogs are each given 12.5

Sergeeva-Olga [200]

12*12.5=150 oz 18*5=90 oz 150+90=240 convert to pounds = 15 pounds

8 0

3 years ago

Elizabeth wants to ride her bicycle 28.6 miles this week. She has already ridden 19 miles. If she rides for 3 more days, write a

8090 [49]

Answer: The average number of miles she would have to ride each day to meet her goal = 3.2 miles.

Step-by-step explanation:

Given: Elizabeth wants to ride her bicycle 28.6 miles this week.

She already ridden 19 miles.

Number of days she will ride more =3

Let m = average number of miles ridden in a day.

Then,

Total distance = initial distance + (m) (Number of days)

i.e. 28.6 = 19+3m (Required equation)

Subtract 19 from both sides , we get

$9.6=3m\\\\\Rightarrow\ m=\dfrac{9.6}{3}\\\\\Rightarrow\ m=3.2\ miles$

Hence, the average number of miles she would have to ride each day to meet her goal = 3.2 miles.

5 0

3 years ago