14 sandwiches and 3 drinks cost €33.75. 6 sandwiches and 17 drinks cost €26.25 . How much cost 1 sandwich and 1 drink

suppose that a number written as a decimal has an infinite number of non- repeating digits after the decimal point. What is this

faltersainse [42]

Answer: Irrational number

If the decimal digits repeat forever, then the repeating decimal is considered rational.

For instance, 2/99 = 0.020202020202... where the "02" repeats forever

If we don't have such a pattern, then we cannot write the decimal as a fraction of two integers and the number is not rational. So it is irrational.

8 0

3 years ago

Find the value.plzzz I will mark as the brillianest plzzzzzzzzzzzzzzzzzzzz

Aleks [24]

Answer:

n= 75

y= 75

z= 50

p= 105

other n across from p= 105

Step-by-step explanation:

n and y are the same since the triangle is isoseles.

p is supplemetary to n

z and 50 are opposites in a quadrilateral

7 0

3 years ago

Consider the sets below. u = {x | x is a real number} a = {x | x is an odd integer} r = {x | x = 3, 7, 11, 27} is r ⊂ a?

-BARSIC- [3]

The correct option is (B) yes because all the elements of set R are in set A.

<h3>What is an element?</h3>

In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.

Given sets:

U = {x | x is a real number}
A = {x | x is an odd integer}
R = {x | x = 3, 7, 11, 27}

So,

A = 1, 3, 5, 7, 9, 11... are the elements of set A.
R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.

Therefore, the correct option is (B) yes because all the elements of set R are in set A.

Know more about sets here:

brainly.com/question/2166579

#SPJ4

The complete question is given below:
Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A?

(A) yes, because all the elements of set A are in set R

(B) yes, because all the elements of set R are in set A

(C) no because each element in set A is not represented in set R

(D) no, because each element in set R is not represented in set A

8 0

1 year ago

Which angle has a sine of -1/2 and a cosine of -√3/2

Sergeeva-Olga [200]

$\sin x=-\dfrac{1}{2} < 0\\\\\cos x=-\dfrac{\sqrt3}{2} < 0\\\\therefore\ x\in(180^o;\ 270^o)$

$\text{We know:}\ \tan x=\dfrac{\sin x}{\cos x}\\\\\text{therefore:}\ \tan x=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt3}=\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{\sqrt3}{3}$

$\tan x=\dfrac{\sqrt3}{3}\Rightarrow x=30^o+180^o\cdot k;\ k\in\mathbb{Z}$

$x\in(180^o;\ 270^o)\ therefore\ \boxed{x=30^o+180^o=210^o}$

8 0

3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago