2x minus negative 4x

Which number is less than -2 4/5 but greater than -3 1/5

Alex_Xolod [135]

If you need help with things like this, put the fractions into decimals.
A number that is less than -2.8 and greater than -3.2

A -3.5
B -3.666
C -2.5
D -2.875
The answer is C

5 0

3 years ago

Simplify: 11y-110/50-5y

alexandr1967 [171]

<h2>Answer:</h2> $\frac{11y-110}{50-5y}=-\frac{11}{5}$

<h2>Explanations:</h2>

Given the following expression show below;

$\frac{11y-110}{50-5y}$

Factor out the GCF from both the numerator and denominator

$\frac{11(y-10)}{5(10-y)}$

This can also be expressed as;

$\begin{gathered} \frac{11(-10+y)}{5(10-y)} \\ \frac{-11(10-y)}{5(10-y)} \end{gathered}$

Cancel out the common expression

$\frac{-11\cancel{(10-y)}}{5\cancel{(10-y)}}=-\frac{11}{5}$

Hence the simplified form of the expression is -11/5

7 0

1 year ago

Find the largest interval which includes x = 0 for which the given initial-value problem has a unique solution. (Enter your answ

Ivan

Answer:

$(-\infty,3)$

Step-by-step explanation:

We are given that

$(x-3)y''+4y=x$

$y''+\frac{4}{x-3}y=\frac{x}{x-3}$

y(0)=0

y'(0)=1

By comparing with

$y''+p(x)y'+q(x)y=g(x)$

We get

$p(x)=\frac{4}{x-3}$

$g(x)=\frac{x}{x-3}$

q(x)=0

p(x),q(x) and g(x) are continuous for all real values of x except 3.

Interval on which p(x),q(x) and g(x) are continuous

$(-\infty,3)$ and (3, $\infty)$

By unique existence theorem

Largest interval which contains 0= $(-\infty,3)$

Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution= $(-\infty,3)$

7 0

3 years ago

Which of the following graphs could be the graph of the function f(x)=-0.08x(x^2-11x+18)?

MariettaO [177]

Answer:

option C

Step-by-step explanation:

f(x)=-0.08x(x^2-11x+18)

Factor the parenthesis and find out the x intercepts

(x^2-11x+18)

We find out two factors whose product is 18 and sum is -11

-9 and -2 gives us product 18

-9 -2 gives us sum -11

(x^2-11x+18) =(x-9)(x-2)

f(x)=-0.08x(x^2-11x+18)

f(x)= -0.08x(x-9)(x-2)

set f(x) =0 and solve for x

-0.08x(x-9)(x-2) =0

-0.08x =0, so x=0

x-9 = 0 , so x=9

x-2 =0 , so x=2

x intercepts are (0,0) (2,0) and (9,0)

x intercepts are not repeating

so the graph crosses x -axis at x=0, 2,9

option C is correct

The graph is attached below4

5 0

3 years ago

Read 2 more answers

Problema de capacidades Una empresa aceitera ha envasado 3000 litros de aceite en 1200 botellas de dos y de cinco litros. ¿Cuánt

mariarad [96]

Answer:

Hay 200 botellas de 5 litros y 1000 botellas de 2 litros.

Step-by-step explanation:

Un sistema de ecuaciones lineales es un conjunto de dos o más ecuaciones de primer grado, en el cual se relacionan dos o más incógnitas.

Resolver un sistema de ecuaciones consiste en encontrar el valor de cada incógnita para que se cumplan todas las ecuaciones del sistema.

En este caso, las variables a calcular son:

x= cantidad de botellas de 2 litros.
y= cantidad de botellas de 5 litros.

Una empresa aceitera ha envasado 3000 litros de aceite en 1200 botellas de dos y de cinco litros. Entonces es posible plantear el siguiente sistema de ecuaciones:

$\left \{ {{2*x+5*y=3000} \atop {x+y=1200}} \right.$

Existen varios métodos para resolver un sistema de ecuaciones. Resolviendo por el método de sustitución, que consiste en despejar o aislar una de las incógnitas y sustituir su expresión en la otra ecuación, despejas x de la segunda ecuación:

x= 1200 - y

Sustituyendo la expresión en la primer ecuación:

2*(1200 - y) + 5*y=3000

Resolviendo se obtiene:

2*1200 - 2*y + 5*y= 3000

2400 +3*y= 3000

3*y= 3000 - 2400

3*y= 600

y= 600÷3

y= 200

Reemplazando en la expresión x= 1200 - y:

x= 1200 - y

x=1200 -200

x= 1000

<u><em>Hay 200 botellas de 5 litros y 1000 botellas de 2 litros.</em></u>

7 0

2 years ago