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2 years ago

Least common multiple of 6 and 8

Mathematics

2 answers:

kicyunya [14]2 years ago

7 0

$\huge\text{Hey there!}$

$\large\textsf{What are the least common multiples of 6 and 8?}$

$\large\textsf{Well, first lets find the multiples of both of the numbers \& see}\\\large\textsf{what we could find.}$

$\huge\textbf{The MULTIPLES of 6}\\\bullet\large\text{ 6, 12, 18,24, 30, and 36}$

$\huge\textbf{The MULTIPLES of 8}\\\bullet\large\text{ 8, 16, 24, 32, and 40}$

$\large\textsf{Each of the numbers listed above shared ONE number \& it is 24}$

$\huge\text{Answer: \boxed{\mathsf{\bold{L}owest \ \bold{C}ommon \ \bold{M}ultiple: 24}}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

miskamm [114]2 years ago

3 0

Answer:

8={8,16,24,32,40}

6={6,12,18,24,33}

so the least common multiple is 24

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Answer:

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Step-by-step explanation:

6r = 95-47

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r = 48/6

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6. A two-story building measures 21 feet

Y_Kistochka [10]

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Step-by-step explanation:

First figure the ratio of the real building to the model which is 21 divide 3. So the ratio is 1:7, since we're finding the height of the model, we divide 7 by 84 which is 12

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Please help!<br><br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7B4.2%7D%7B7%5Cfrac%7B5%7D%7B7%7D%20%7D%20%3D%20%20%5Cfrac%7B3%5C

emmasim [6.3K]

Answer:

d = 40/7

Step-by-step explanation:

Solve for d:

0.544444 = 28/(9 d)

0.544444 = 49/90:

49/90 = 28/(9 d)

49/90 = 28/(9 d) is equivalent to 28/(9 d) = 49/90:

28/(9 d) = 49/90

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3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

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Andru [333]

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