All categories

3 years ago

What is the percent uncertainty in the measurement 2.58 ± 0.15 cm?

Mathematics

1 answer:

saveliy_v [14]3 years ago

8 0

$\frac{0.15}{2.58} \times 100\% = 5.814\%$

You might be interested in

Find the slope. Name the unit rate

jok3333 [9.3K]

Answer:

The slope is 4/5 i dont know unit rate though, hope this helps:)

8 0

2 years ago

(05.06)Keisha ran to swim practice but needed to stop by her friend Natasha’s house on the way. She stayed at Natasha’s house fo

Ugo [173]

The graph third represents the provided situation first line increases and then constant then again increases and becomes constant.

<h3>What is a line graph?</h3>

A line graph is a graph made up of segments of straight lines that connect the depicted data points.

We have:

Keisha ran to swim practice but needed to stop by her friend Natasha’s house on the way. She stayed at Natasha’s house for 15 minutes, and together they ran the rest of the way to practice, where they stayed for 1 hour.

Based on the information given we can say graph third represents the situation.

First Keisha ran to swim practice so the line first increases and then she but needed to stop by her friend Natasha’s house on the way. She stayed at Natasha’s house for 15 minutes which means the line should be constant.

Again they both ran so line increases.

Then they both stayed for 1 hour, so the line should be constant.

Thus, the graph third represents the provided situation first line increases and then constant then again increases and becomes constant.

Learn more about the line chart here:

brainly.com/question/23680294

#SPJ1

7 0

2 years ago

Solve 2 log x = log 36<br> Answer<br> x = 72<br> x = 6<br> x = ±6<br> x = 18

astra-53 [7]

$2\log x=\log 36\\
\log x^2=\log 36\\
x^2=36\\
x=-6 \vee x=6
$

6 0

3 years ago

Read 2 more answers

A mass weighing 16 pounds stretches a spring (8/3) feet. The mass is initially released from rest from a point 2 feet below the

mezya [45]

Answer with Step-by-step explanation:

Let a mass weighing 16 pounds stretches a spring $\frac{8}{3}$ feet.

Mass= $m=\frac{W}{g}$

Mass= $m=\frac{16}{32}$

$g=32 ft/s^2$

Mass,m= $\frac{1}{2}$ Slug

By hook's law

$w=kx$

$16=\frac{8}{3} k$

$k=\frac{16\times 3}{8}=6 lb/ft$

$f(t)=10cos(3t)$

A damping force is numerically equal to 1/2 the instantaneous velocity

$\beta=\frac{1}{2}$

Equation of motion :

$m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)$

Using this equation

$\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)$

$\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)$

$\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)$

Auxillary equation

$m^2+m+12=0$

$m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}$

$m=\frac{-1\pmi\sqrt{47}}{2}$

$m_1=\frac{-1+i\sqrt{47}}{2}$

$m_2=\frac{-1-i\sqrt{47}}{2}$

Complementary function

$e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})$

To find the particular solution using undetermined coefficient method

$x_p(t)=Acos(3t)+Bsin(3t)$

$x'_p(t)=-3Asin(3t)+3Bcos(3t)$

$x''_p(t)=-9Acos(3t)-9sin(3t)$

This solution satisfied the equation therefore, substitute the values in the differential equation

$-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)$

$(3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)$

Comparing on both sides

$3B+3A=20$

$3B-3A=0$

Adding both equation then, we get

$6B=20$

$B=\frac{20}{6}=\frac{10}{3}$

Substitute the value of B in any equation

$3A+10=20$

$3A=20-10=10$

$A=\frac{10}{3}$

Particular solution, $x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

Now, the general solution

$x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

From initial condition

x(0)=2 ft

x'(0)=0

Substitute the values t=0 and x(0)=2

$2=c_1+\frac{10}{3}$

$2-\frac{10}{3}=c_1$

$c_1=\frac{-4}{3}$

$x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)$

Substitute x'(0)=0

$0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2$

$\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0$

$\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}$

$c_2==-\frac{64}{3\sqrt{47}}$

Substitute the values then we get

$x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

8 0

3 years ago

In a quiz , team A scored 2 ,4 , -2 , 0 , -2 team B scored -4 ,2 , 2 ,-2 ,4 in successive5 rounds

Eduardwww [97]

Answer:

a. 2 b. 2

Step-by-step explanation:

8 0

3 years ago