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3 years ago

What is the common ratio of the sequence?

1 answer:

6 0

Hello!

To find this you have to divide the number by its previous one

6 / -2 = -3
-18 / 6 = -3
54/-18 = -3

The answer is A) -3

Hope this helps!

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I need help with these three problems.... please help me????

viva [34]

12. Answer: 2130

Step-by-step explanation:

T = -0.01y + 10.6

9.2 = -0.01y + 10.6

-10.6 -10.6

-1.4 = -0.01y

÷-0.01 ÷-0.01

140 = y

"y" represents the number of years after 1990

1990 + 140

= 2130

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13. Answer: -0.0036° per foot

Step-by-step explanation:

Since the relationship is linear, then we can set up the values as x, y coordinates and use the slope formula to find the rate of degrees over feet.

(6562, 63) and (16405, 28)

rate = $\frac{28 - 63}{16405-6562}$

= $\frac{-35}{9843}$

= $\frac{-0.0036}{1}$

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14. Answer: $0.28 per kWh

Step-by-step explanation:

Same reason as #13

(200, 55) and (500, 140)

rate = $\frac{140-55}{500-200}$

= $\frac{85}{300}$

= $\frac{0.28}{1}$

5 0

3 years ago

Which expression can be used to detemine the reference for the angle, x, measuring 150 degree

taurus [48]

Do you have an A B C it helps

4 0

3 years ago

If it’s 3:00 how long will it take before the hands are at a right angle again

UNO [17]

Answer:

4 hours

Step-by-step explanation:

3+1=4

3 0

3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago

Two-step equations - integers

Andrews [41]

Question 1. -3 + v5 = 0
Work : -3 + 5v = 0
5v = 3 ( divide )
answer v = 3/5

Question 2. 1 - 7B = -20
work : -7b = -20 - 1
-7b = -21 ( divide )
answer b = 3

question 3 : -k - 5 = 0
work : -k = 5
answer : k = - 5

Question 4 : -1 + 8a = 129
work : 8a = 129 + 1
8a = 130 (divide )
answer a = 65/4

3 0

3 years ago

Read 2 more answers