All categories

3 years ago

Help please on math plz

Mathematics

1 answer:

Rina8888 [55]3 years ago

7 0

it is b because it has no numbers that repeat

You might be interested in

If you could I would appreciate it if you can subscribe to my YouTub/e channel called 2k Vibes then click filter then channel th

kow [346]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Original price: $35 What’s the percent of the discount Sale price: $31.50

Harrizon [31]

Answer:

10%

Step-by-step explanation:

You can solve this equation by back-solving. Pick a random percent that you would think is close and convert it to decimal, then multiply it by the original price and subtract that from the original price.

5 0

3 years ago

Pls helppp and tap on the image to see it

musickatia [10]

Answer:

See attached

Step-by-step explanation:

The midpoint will be -> (3, 2)

We get this from plugging in our numbers (file 2) and solving. You get (6/2, 4/2) which results in (3, 2).

Files:

[1] Midpoint Formula

[2] I filled in the image you shared

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

4 0

3 years ago

1+10= ? 10+1= ? 10 + 5 = ? LoL

mina [271]

Answer:

2758.653

Step-by-step explanation:

I'm super smart

7 0

3 years ago

: Show that the solution of the differential equation: = − − − − − is of the form: + + ( − ) = + , When = and =

Serhud [2]

Answer:

$y = \tan(x + \frac{x^2}{2})$

Step-by-step explanation:

Poorly formatted question; The complete question requires that we prove that $y=\tan(x+\frac{x\²}{2})$

When

$\frac{dy}{dx} =1+xy\²+x+y\²$ and $y(0)=0$

We have:

$\frac{dy}{dx} =1+xy\²+x+y\²$

Rewrite as:

$\frac{dy}{dx} =1+x+xy\²+y\²$

Factorize

$\frac{dy}{dx} = (1+x)+y\²(x+1)$

Rewrite as:

$\frac{dy}{dx} = (1+x)+y\²(1+x)$

Factor out 1 + x

$\frac{dy}{dx} = (1+y\²)(1+x)$

Multiply both sides by $\frac{dx}{1 + y^2}$

$\frac{dy}{1+y\²} = (1+x)dx$

Integrate both sides

$\int \frac{dy}{1+y\²} = \int (1+x)dx$

Rewrite as:

$\int \frac{1}{1+y\²} dy = \int (1+x)dx$

Integrate the left-hand side

$\int \frac{1}{1+y\²} dy = \tan^{-1}y$

Integrate the right-hand side

$\tan^{-1}y = x + \frac{x^2}{2} + c$

$y(0)=0$ implies that: $(x,y) = (0,0)$

So:

$\tan^{-1}y = x + \frac{x^2}{2} + c$ becomes

$\tan^{-1}(0) = 0 + \frac{0^2}{2} + c$

This gives:

$0 = 0 +0 + c$

$0 =c$

$c = 0$

The equation $\tan^{-1}y = x + \frac{x^2}{2} + c$ becomes

$\tan^{-1}y = x + \frac{x^2}{2} + 0$

$\tan^{-1}y = x + \frac{x^2}{2}$

Take tan of both sides

$y = \tan(x + \frac{x^2}{2})$ --- Proved

8 0

3 years ago