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
Readme [11.4K]
2 years ago
7

Which ones are variable expenses? (Check all that apply)

Mathematics
1 answer:
mr Goodwill [35]2 years ago
7 0

Answer:

food bill, electric bill

Step-by-step explanation:

Variable expenses change every money

Loan payments are usually fixed, the same every month

food bill - variable expense since you buy different food each month and they will cost different amounts

rent is a fixed expense

the electric bill is different every month because some months you use more electricity than others

You might be interested in
Solve the system of equations in the workspace below: <br> 2x - y = 16<br> x + y = 5
Ad libitum [116K]
Already you can see that -y and y will make a zero
add both together
3x=21
x=7

plug x in
7+y=5
y=-2
7 0
3 years ago
A) Compute the sum
avanturin [10]
A)

To calculate this sum, we could use trigonometric identity:

\arcsin(x)-\arcsin(y)=\arcsin\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)

We have:

\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k+1-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{(k+1)^2-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\&#10;

=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\dfrac{\sqrt{(k+1)^2-1}}{\sqrt{(k+1)^2}}-\dfrac{1}{k+1}\cdot\dfrac{\sqrt{k^2-1}}{\sqrt{k^2}}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{\dfrac{(k+1)&^2-1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{\dfrac{k^2-1}{k^2}}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\dfrac{1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{1-\dfrac{1}{k^2}}\right]=\\\\\\&#10;

=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\left(\dfrac{1}{k+1}\right)^2}-\dfrac{1}{k+1}\cdot\sqrt{1-\left(\dfrac{1}{k}\right)^2}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\left[\arcsin\left(\dfrac{1}{k}\right)-\arcsin\left(\dfrac{1}{k+1}\right)\right]=\\\\\\

=\bigg[\arcsin(1)-\arcsin\left(\frac{1}{2}\right)\bigg]+\bigg[\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)\bigg]+\\\\\\+&#10;\bigg[\arcsin\left(\frac{1}{3}\right)-\arcsin\left(\frac{1}{4}\right)\bigg]+\dots+&#10;\bigg[\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)\bigg]=\\\\\\

=\arcsin(1)-\arcsin\left(\frac{1}{2}\right)+\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{1}{3}\right)-\\\\\\-\arcsin\left(\frac{1}{4}\right)+\dots+\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)=\\\\\\=&#10;\arcsin(1)-\arcsin\left(\frac{1}{n+1}\right)=\dfrac{\pi}{2}-\arcsin\left(\frac{1}{n+1}\right)

So the answer is:

\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)}

B)

\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\lim\limits_{n\to\infty}\Bigg(\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)\Bigg)=\dfrac{\pi}{2}-\lim\limits_{n\to\infty}\arcsin\left(\dfrac{1}{n+1}\right)=\\\\\\=&#10;\Bigg\{\dfrac{1}{n+1}\xrightarrow{n\to\infty}0\Bigg\}=\dfrac{\pi}{2}-\arcsin(0)=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}

So we prove that:

\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}
7 0
3 years ago
Help help help help help
alexdok [17]

Answer:

\sf 5x+6y=12

Step-by-step explanation:

\sf y=-\dfrac{5x}{6}+2

Multiply both sides by 6:

\implies \sf 6 \cdot y=6(-\dfrac{5x}{6}+2)

Expand:

\implies \sf 6y=-5x+12

Add 5x to both sides:

\implies \sf 5x+6y=-5x+12+5x

\implies \sf 5x+6y=12

6 0
2 years ago
Need help please 9× &gt; 108​
butalik [34]

Answer:

x > 12

Step-by-step explanation:

9x>108

x> 108/9

x > 12

8 0
3 years ago
Read 2 more answers
Find an equation of a line parallel to y=-x-2 and passes through the point (2,-2). Write answer form y=mx+b
tangare [24]
To be parallel, this new line has to have the same slope as the other. To find the slope, we look at the number multiplying x.

-x means x is being multiplied by -1, so the slope, or m, is -1.

y = -x + b

Now, input the coordinate point and solve for b.

-2 = -1(2) + b
-2 = -2 + b
Add 2 to both sides.
0 = b

y = -x
5 0
4 years ago
Read 2 more answers
Other questions:
  • Math homework help please thanks!
    7·1 answer
  • What is the reciproncal of fraction 2 5th
    8·2 answers
  • Answer for number 2.That's inequality. .I already got the answer but I don't know how to do it..Plz show your step that how you
    10·1 answer
  • an animal shelter in a 36 kittens and 12 puppies available for adoption what is the ratio of puppies to kittens
    10·1 answer
  • What 2 +2 please someone tell me​
    5·1 answer
  • What is the volume of a box that is 9 cm long. 6 cm<br>wide, and 4 cm high?<br><br>​
    14·2 answers
  • Find the perimeter of the shape below:
    15·2 answers
  • A die is rolled. The set of equally likely outcomes is {1, 2, 3, 4, 5, 6}. Find the probability of getting a 9.
    12·1 answer
  • While attending a film festival you decide that there are 13 movies that you are interested in seeing. However, you only have ti
    14·2 answers
  • The Shoe Outlet bought boots for $60 and marks up the boots 55% on the selling price. What is the selling price of the boots?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!