1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
blsea [12.9K]
3 years ago
14

Fractions on a number line.

Mathematics
1 answer:
larisa [96]3 years ago
8 0
1/5  3/5  7/10  4/5  Change the denominators all to 10 to  be able to but them in order
You might be interested in
Lydia baked a total of 144 chocolate chip cookies and peanut butter treats for Valentine's Day. Initially, the ratio of chocolat
andre [41]
Let x be the chocolate chip cookies, so the peanut butter treats will be 144-x.
We know that the cookies and the treats are in a ratio of 5:3, so:\frac{cookies}{treats} = \frac{5}{3} = \frac{x}{144-x}
Now we can solve for x:
\frac{5}{3} = \frac{x}{144-x}
5(144-x)=3x
720-5x=3x
8x=720
x= \frac{720}{8}
x=90

We now know Lydia has 90 chocolate chip cookies, and 144-x=144-90=54 peanut butter treats.

Then, Lydia's friend ate \frac{3}{5} of her cookies, so her friend ate \frac{2}{5} (90)=36 cookies. Therefore, Lydia has 90-36=54 cookies left.
Now we can calculate the total amount of baked goods after her friend ate the cookies:
144-54=90
Therefore, our remainder treats will be:
90-x

We also now that after her friend ate 54 cookies and some treats, the new ratio is 6:1, and that's all we need to set up our new equation and solve it to find how many treats she ate:
\frac{cookies}{treats} = \frac{6}{1} = \frac{54}{90-x}
6(90-x)=54
540-6x=54
6x=486
x= \frac{486}{6}
x=81

Finally, if she ate 81 out 90 treats, we can conclude the Lydia has left with 9 peanut butter treats.


3 0
4 years ago
Read 2 more answers
Find the value of the expression 0.5^10/0.5^7
SIZIF [17.4K]

Answer:

The solution is:

  • \frac{0.5^{10}}{0.5^7}=0.125

Step-by-step explanation:

Given the expression

0.5^{10}\div 0.5^7

Solving

\frac{0.5^{10}}{0.5^7}

\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}

so

\frac{0.5^{10}}{0.5^7}=0.5^{10-7}

\mathrm{Subtract\:the\:numbers:}\:10-7=3

=0.5^3

=0.125        ∵ 0.5^3=0.125

Therefore, the solution is:

\frac{0.5^{10}}{0.5^7}=0.125

7 0
3 years ago
The general form of the equation of a circle is Ax2 + By2 + Cx + Dy + E = 0, where A = B 0. If the circle has a radius of 3 unit
Anettt [7]
(x-h)^2 + (y-k)^2 = r^2

r=3
(x-h)^2 + (y-k)^2 = 3^2

the center lies on the y-axis --> h=0
x^2 + (y-k)^2 = 3^2 = 9

expand
x^2 + y^2 -2ky + k^2 = 9

x^2 + y^2 -2ky + (k^2 - 9) = 0
compare to general form
A=1 , B=1 and C =0
D= -2k and E=k^2 - 9
8 0
3 years ago
Read 2 more answers
Help me out. good points the image shows​
Sphinxa [80]
To start off with you need to get math-way it will help out
5 0
3 years ago
Read 2 more answers
roblem: Report Error Let $f(x)=(x^2+6x+9)^{50}-4x+3$, and let $r_1,r_2,\ldots,r_{100}$ be the roots of $f(x)$. Compute $(r_1+3)^
Tasya [4]

Answer:

9

Step-by-step explanation:

you take the L bud

3 0
4 years ago
Other questions:
  • You walk 2 miles from your house to the park and 4.5 miles from the park to the lake. Then you return home along a straight path
    5·2 answers
  • The sum of two numbers is 121. If the larger number is thirteen more than the smaller number, find both numbers
    14·1 answer
  • A pair of intersecting planes and where they intersect
    8·1 answer
  • .83 times 4.52 I need work
    15·1 answer
  • Calcule o valor de X<br>3,5+×=7,5​
    12·1 answer
  • Find the other endpoint (E2) given the midpoint (2,2) and<br> endpoint 1 (-3,-2)
    8·1 answer
  • The total population of an economy is 175 million, the labor force is 125 million, and the number of employed workers is 117 mil
    11·1 answer
  • What is the solution set of the compound inequality
    10·1 answer
  • Please, can you help.... I'm really confused. I need an answer really fast. Thank you
    7·1 answer
  • Help!!
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!