All categories

3 years ago

Which rational number equals 0 point 3 with bar over 3?

Mathematics

2 answers:

Sergio039 [100]3 years ago

7 0

Answer:

0.333.. repeating = $\frac{1}{3}$

Step-by-step explanation:

The given number is 0.333...

The bar represents 3 continues forever.

Let's find the rational number that represents 0.333...

The rational number is in the form of $\frac{p}{q}$ , where q ≠ 0.

Let x = 0.333...

Let's multiply both sides by 10, we get

10x = 3.33...

Now let's subtract x from 10x

10x - x = 3.33... - 0.333..

9x = 3

Now let's divide both sides by 9, we get

$\frac{9x}{9} = \frac{3}{9}$

x = $\frac{3}{9}$

When we simplify the above, we get

x = $\frac{1}{3}$

So the rational number $\frac{1}{3}$ represents 0.333....

Juli2301 [7.4K]3 years ago

3 0

The overbar indicates the digit repeats forever. It is worth remembering that

$0.\overline{x}=\dfrac{x}{9}\qquad\text{x = any single digit}$

Your number is 3/9 = 1/3

You might be interested in

Students are planting a flower bed for science class. The flower bed can be 4.5 ft wide and will be divided into 2 sections for

aksik [14]

Answer:

3(4.5) Because they equal the same thing

Step-by-step explanation:

4.5(+3)=13.5

3(4.5)=13.5

7 0

2 years ago

This question has three parts. Answer the parts in order.

nalin [4]

Answer:

The area of the smallest section is $A_{1}=100yd^{2}$

The area of the largest section is $A_{2}=625yd^{2}$

The area of the remaining section is $A_{3}=250yd^{2}$

Step-by-step explanation:

Please see the picture below.

1. First we are going to name the side of the larger square as x.

As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:

- Area of the first section:

$A_{1}=10yd*10yd$

$A_{1}=100yd^{2}$

- Area of the second section:

$A_{2}=x^{2}$ (Eq.1)

- Area of the third section:

$A_{3}=width*length$

$A_{3}=10yd*x$ (Eq.2)

2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:

$A_{1}+A_{2}+A_{3}=975yd^{2}$

Replacing values:

$100+x^{2}+10x=975$

Solving for x:

$x^{2}+10x-875=0$

$x=\frac{-10+\sqrt{100+(4*875)}}{2}$

$x=\frac{-10+\sqrt{3600}}{2}$

$x=\frac{-10+60}{2}$

$x=25$

3. Replacing the value of x in Eq.1 and Eq.2:

- From Eq.1:

$A_{2}=25^{2}$

$A_{2}=625yd^{2}$

- From Eq.2:

$A_{3}=10*25$

$A_{3}=250yd^{2}$

3 0

2 years ago

1. The student council at Camp Creek Middle School determines that 3 out of 4

blagie [28]

Answer:

Hi there!

Your answer is: 150 people prefer that school assemblies be held on Friday afternoon

Step-by-step explanation:

3 out of 4 students means that 75% of students prefer assemblies on Friday afternoon!

To find 75% of 200:

200 = 100%

/100 to find 1%

2 people = 1%

Multiply it by 75 to determine 75%

2×75= 150

150people is 75% of 200 people!

I hope this makes sense! Let me know if you have questions!

7 0

3 years ago

Read 2 more answers

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Which equations may be used to determine the 3 consecutive integers whose sum is 127?

s344n2d4d5 [400]

Answer:

a and b.

Step-by-step explanation:

I’ll explain by giving an example.

Let’s say that: a=3;b=4;c=5; => they all are consecutive -> their sum is 12.

=> if we use a) n=3 => 3*n+3=3*3+3=12 => correct.

b) n+(n+1)+(n+2)= 3+4+5=12=> correct.

c)n+2n+3n=3+6+9=18=>incorrect.

d)3n=3*3=9=>incorrect.

7 0

2 years ago