All categories

3 years ago

If your profit in a deal was 15% and you got rupees 120 the deal was worth

2 answers:

8 0

Percentage of profit made on the deal = 15%
Amount of profit made on the deal = rupees 120
Let us assume the deal was worth the amount = x
Then we can get the equation as
15x/100 = 120
15x = 120 * 100
15x = 12000
x = 12000/15
= 800 rupees
So the deal was worth 800 rupees. Most of the required information’s are already given in the question. Only thing is using those information’s to get to the required answer.

MrRissso [65]3 years ago

5 0

If you would like to know how much was the deal worth, you can calculate this using the following steps:

15% of the deal is 120
15% * x = 120
15/100 * x = 120 /*100/15
x = 120 * 100 / 15
x = 800

Result: The deal was worth 800.

You might be interested in

Solve each system by elimination. 3x - 2y = 1 2x + 2 y = 14

Paha777 [63]

Add 3 and 12 , cancel -2 y+2y and and add 14 and 14
3x-2y=14
12x+2y=14
15x=28
Divide both sides by 15
X=28/15
3x 28/15 -2y=14
Y=-21/5
(X,y)=28/15, -21/5
3x 28/15 -2x(-21/5)=12x 28/15 +2x(-21/5)=14
14=14=14
(x, y)=(28/15, -21/5) is the answer

6 0

3 years ago

What type of triangle is △STR? right triangle equilateral triangle isosceles triangle scalene triangle

Setler79 [48]

Answer:

C. Isosceles Triangle

Step-by-step explanation:

An Isosceles triangle is a type of triangle in geometry that has two sides of equal length.

From the figure attached, triangle STR has two sides of equal length which are;

1.) ST and

2.) TR

The two equal sides (ST and TR) are called legs while the third side is called the base of the triangle.

*P.S: I believe your question is based on the image attached

7 0

2 years ago

Someone please help me !

Greeley [361]

(2,5) I think I’m sorry if I’m wrong:/

5 0

3 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago

Yoxel buys 4 1/2 gallons of soda. 1/4of the soda he bought was Pepsi and the rest was Sprite. How many gallons of Pepsi did Yoxe

marusya05 [52]

Answer:

$4\frac{1}{4}$ gallons