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alexgriva [62]
3 years ago
7

A runner ran 1 4/5 miles on Monday and 6 3/10 on Tuesday. How many times her Monday's distance was her Tuesday's distance? Pleas

e help me I need an answer that makes sense
Mathematics
1 answer:
Tamiku [17]3 years ago
8 0
He ran more than enough that’s all extra talk
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Answer:Y axis

Step-by-step explanation:

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Using your equation from #2, how many hours did the plumber work at your home? Solve the equation from #2
jek_recluse [69]

Answer:

the answer i got was 3 which would be none of the above bc i minused 40 on both sides to get 150 then i divided 150/50 which got me 3.

Step-by-step explanation:

:) your welcome

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3 years ago
Read 2 more answers
Find each of the following for ​
KATRIN_1 [288]
<h2>Answer:</h2>

(a)

          f(x+ h)=8x+8h+3  

(b)

            f(x+ h)-f(x)=8h          

(c)

             \dfrac{f(x+ h)-f(x)}{h}=8

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as :

              f(x)=8x+3

(a)

           f(x+ h)

We will substitute (x+h) in place of x in the function f(x) as follows:

f(x+h)=8(x+h)+3\\\\i.e.\\\\f(x+h)=8x+8h+3

(b)

       f(x+ h)-f(x)              

Now on subtracting the f(x+h) obtained in part (a) with the function f(x) we have:

f(x+h)-f(x)=8x+8h+3-(8x+3)\\\\i.e.\\\\f(x+h)-f(x)=8x+8h+3-8x-3\\\\i.e.\\\\f(x+h)-f(x)=8h

(c)

           \dfrac{f(x+ h)-f(x)}{h}            

In this part we will divide the numerator expression which is obtained in part (b) by h to get:

           \dfrac{f(x+ h)-f(x)}{h}=\dfrac{8h}{h}\\\\i.e.\\\\\dfrac{f(x+h)-f(x)}{h}=8    

5 0
3 years ago
A graph that has a finite or limited number of data points is
Mandarinka [93]

a discrete graph (I had to look this one up)

4 0
3 years ago
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
3 years ago
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