All categories

2 years ago

56+24x product of the gcf and sum

Mathematics

1 answer:

sergejj [24]2 years ago

3 0

SOLUTION:

GCF = 8

56 + 24x = 8 ( 7 + 3x )

Hope this helps! :)

You might be interested in

Find the quotient 400÷8

lyudmila [28]

The quotient is 50 :)

3 0

2 years ago

Read 2 more answers

Jesus loves you repent your sins because he is coming back

miv72 [106K]

thats cool did you want to ask a question though?

7 0

3 years ago

Read 2 more answers

How do you solve 3 1/4- 1 7/8 step by step?

Vilka [71]

Answer:

1 3/8.

Step-by-step explanation:

3 1/4 - 1 7/8

First convert to improper fractions:

3 1/4 = (3*4) + 1 / 4 = 13/4.

1 7/8 = (1 *8)+7 / 8 = 15/8. So we have:

13/4 - 15/8

The Lowest Common Multiple of 4 and 8 is 8 so we convert 13/4 to eighth's

13 / 4 = 13*2 / 4 *2 = 26/8

Finally: we have

26/8 - 15/8 = (26 - 15) /8

= 11 /8

= 1 3/8 (answer)

4 0

3 years ago

BRAINLIEST + 7 POINTS TO FIRST CORRECT ANSWER

timama [110]

Answer:

it has 11 faces 10 sides for the decagon + 1 for the base

It does have one face that is a decagon, which is the base, the rest are triangles

it has 20 edges

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

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