Ask question

Ask question

All categories

3 years ago

10

Eggs ares sold in boxes. A small box holds 6 eggs hina buys x small boxes of eggs write down in terms of x the total number of e

ggs in these small boxes

1 answer:

krok68 [10]3 years ago

5 0

$6x$

............................................

You might be interested in

A jar has 32 marbles. Some are blue and the rest are green. There are three times as many green marbles as blue. How many green

ella [17]

Answer: 24

Step-by-step explanation:

Number of marbles in jar = 32

Let blue marbles be represented by b.

Since there are three times as many green marbles as blue. This means that green will be = 3 × b = 3b

We then add them together. This will be:

3b + b = 32

4b = 32

b = 32/4

b = 8

There are 8 blue marbles

Since green is 3× blue. Therefore,

Green = 8 × 3 = 24 marbles

7 0

2 years ago

In what time will 64000 amount to 68921 at a rate of 5 % per annum comp half yearly

andre [41]

Answer:

Time = T half–years , Rate = 5 % per half year

2

4 0

3 years ago

Read 2 more answers

Can the sum of three consecutive integers be even? Give some examples to support your Answer.

Alexeev081 [22]

1+2+3= 6 (even)
2+3+4= 9 (odd)
it can be both

6 0

3 years ago

Find the final cost of an item that was originally price at $245 with a 15% discount and a 6% tax. Please help this is My last q

Soloha48 [4]

Answer:

12

Step-by-step explanation:

8 0

2 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

2 years ago