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3 years ago

What is the least common multiple of 4 and 10?

Mathematics

2 answers:

Alexus [3.1K]3 years ago

6 0

Answer:

Step-by-step explanation:

olganol [36]3 years ago

3 0

Answer:

Step-by-step explanation:

Least common multiple (LCM) of 4 and 10 is 20.

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Esther. Vida and Clair were asked to consider two different cash flows: GH¢1000 that theycould receive today and GH¢3000 that wo

xxMikexx [17]

Answer:

The right choice is Esther's

Step-by-step explanation:

Present worth of GH¢ after 3 years is better as its present really worth may be very high than GH¢ a thousand of today. Here fee of interest is not given to compare however GH¢ 3000 after three years will be equivalent to present well worth GH¢ 1000 if charge of interest >44.5% which may be very very excessive. Thus charge of interest > 44.5% Then GH¢ 3000 payable after 3 years could be equivalent to GH¢ 1000 Hence vide has made the proper preference. If interest price >44.5%.

Then Esther's preference will he right. (thinking about time value of money).

6 0

3 years ago

How do i solve 7 - n = 27

LuckyWell [14K]

add 7 to 27 which equals 34

6 0

3 years ago

Read 2 more answers

What additional information is needed to prove the following triangles congruent by ASA?

lianna [129]

Answer:

angle B = angle C

Step-by-step explanation:

______________

7 0

3 years ago

Read 2 more answers

The approximate volume of a sphere with a diameter of 10 centimeters is

Alexeev081 [22]

Answer: The volume is approximately 523.6 cm³.

Step-by-step explanation:

The formula for the volume of a sphere is V=4/3(3.14)r³

The radius is 5 cm, the you just have to multiply everything together.

5 0

3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago