Help me please, i dont know how to solve

On a buying trip in los angeles, rosaria perez ordered 120 pieces of jewelry: a number of bracelets at $77 each and a number

Digiron [165]

Let, the number of bracelets be x and the number of necklaces be y.

Given, there are 120 pieces of jewellery which are bracelets and necklaces.

So we can write the equation, $x+y = 120$ ....equation 1

Now given the price of bracelet = $7

So, the price of x bracelets = $ $(7)(x)$ =$ $(7x)$

Given, the price of necklace = $13.

So, the price of y necklaces = $ $(13)(y)$ = $ $(13y)$

The total amount for the jewellery given = $ 1080.

We can write the equation as,

$7x + 13y = 1080$ ....equation 2

Now from equation 1, we can write,

$x = 120-y$

By substituting this value of x in equation 2, we will get,

$7(120-y) + 13y = 1080$

$840-7y+13y = 1080$

$840+6y = 1080$

Now we will move 840 to the right side by subtracting it from both sides. We will get,

$840+6y-840 = 1080-840$

$6y = 1080-840$

$6y = 240$

We can get y by moving 6 to the other side by dividing it to both sides. We will get,

$\frac{(6y)}{6} =\frac{240}{6}$

$y =\frac{240}{6}$

$y = 40$

So we have got the number of necklaces = 40

Now we have, $x = 120-40$

$x = 80$

So, the number of bracelets = 80

We have got the required answer here.

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3 years ago

How do I make this into a solid number<br><br><img src="https://tex.z-dn.net/?f=3%20%5Cfrac%7B5%7D%7B6%7D%20-%20%20-%201%5Cfrac%

Effectus [21]

Answer:

Basic Of Algebra. Basics of Algebra cover the simple operation of mathematics like addition, subtraction, multiplication, and division involving both constant as well as variables. For example, x+10 = 0. This introduces an important algebraic concept known as equations. Step-by-step explanation:

6 0

3 years ago

The length of an insect’s wing, drawn at a scale of 50:1, is 15 cm. What is the actual length of the wing?

Mkey [24]

50*15cm=750cm
750cm long

7 0

4 years ago

Read 2 more answers

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$