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3 years ago

The sum of three consecutive even numbers is 48. What is the smallest of these numbers?

Mathematics

2 answers:

Helga [31]3 years ago

7 0

The smallest number is 14

SVETLANKA909090 [29]3 years ago

4 0

Answer: The numbers are 14, 16 and 18.

Step-by-step explanation: First thing to note is that consecutive even numbers have a difference of 2 units between every two terms. That is, if the first number is a, the next would be a + 2, and the next would be a + 4, and so on.

Therefore the three consecutive even numbers shall be a, a + 2, and a + 4.

Our equation now becomes;

a + a + 2 + a + 4 = 48

3a + 6 = 48

Subtract 6 from both sides of the equation

3a = 42

Divide both sides of the equation by 3

a = 14

Therefore, the numbers are 14, 16 and 18.

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Use summation notation to write the series 49 + 54 + 59 +... for 14 terms.

My name is Ann [436]

Answer:

Arithmetic sequence states that a sequence of numbers such that the difference between the consecutive terms is constant.

it is given by: $a_n = a+(n-1)d$ where a is the first term , n is the number of term and d is the common difference.

Given the series: $49 + 54 + 59+ ......$

here, Common difference(d) = 5

First term(a) = 49

by definition we have;

For nth term

$a_n = a+(n-1)d$

$a_n = 49+(n-1)5$

$a_n =49+5n -5$ = 5n + 44

To write the series using summation notation for 14 terms

Summation symbol $'\sum'$

The series for 14th terms is given by;

$\sum_{n=1}^{14}(5n +44)$

5 0

3 years ago

Read 2 more answers

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

iren2701 [21]

Hello there, hope I can help!

I assume you mean $L\left\{\frac{ekt+e-kt}{2}\right\}$
With that, let's begin

$\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \ L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}$
$\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b$
$L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}$
$\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}$

$L\left\{t\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \ L\left\{t\right\}=\frac{1}{s^2}$

$L\left\{\frac{e}{2}\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \ L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \ \frac{e}{2s}$

$\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}$

$\frac{ek}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{ek\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{ek}{2s^2}$

$\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{k\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{k}{2s^2}$

$\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}$

Hope this helps!

3 0

3 years ago

Find an equation for the perpendicular bisector of the line segment whose endpoints are (8,-5)(8,−5) and (4,3)(4,3).

Marysya12 [62]

Answer:

$Correct answer\\y=\frac{1}{2} x-4$