All categories

3 years ago

Write this number showing three significant figures. Remember to round. 8.428

Mathematics

2 answers:

zzz [600]3 years ago

8 0

Answer:

The answer is 8.4!

Step-by-step explanation:

To round 8.428 to the nearest tenth consider the hundredths’ value of 8.428, which is 2 and less than 5. Therefore, the tenths value of 8.428 remains 4.

I hope this helped!

-thatLilWeeb

nikdorinn [45]3 years ago

6 0

Answer:

8.43

Step-by-step explanation:

You might be interested in

Which is the equation of a line that has a slope of 2 and contains the point (0,5)

riadik2000 [5.3K]

$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=2(x-0) \\\\\\ y-5=2x\implies y=2x+5$

7 0

3 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

3 years ago

Pls help me with this one 1,2,3,4

Grace [21]

Answer:

Option 2.

Explanation:

That method is the most random in order to decide what students should take the survey.

5 0

3 years ago

Read 2 more answers

If 21 and 35 are two members of a pythagorean triple, what is the third member?

Firlakuza [10]

Answer:

Step-by-step explanation:

21^2+28^2=35^2

7 0

3 years ago

14 POINTS + Brainliest to first correct answer with explanation and diagram.

zhuklara [117]

So for example we have

1/2 divided by 1/6

What we will do it’s

Leave the first one As it is

Turn the second fraction (the one you want to divide by) upside down

The change the divide to multiply

Multiply the first fraction by that reciprocal

Simplify the fraction (if needed)

So it’s will be like this

1/2 x 6/1 = 6/2

Simplify it

6/2 =3

I hope it’s will help u ✨

5 0

3 years ago