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2 years ago

Is 4/12 greater than 1/6

Mathematics

2 answers:

GaryK [48]2 years ago

8 0

Answer- Yes
Explanation- if you simplify 4/12 it would be 1/3 which would still be greater if simplifying helps.

drek231 [11]2 years ago

3 0

Answer:

Yes, 4/12 is greater than 1/6

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Find the slope between: (1,-2) (2,-2)

stiks02 [169]

Answer: the slope would be 0

Step-by-step explanation:

7 0

2 years ago

What is the slope of the line that passes through (3, -4) and (-2, 5). A)-5/9 B)-9/5 C)9/5 D)5/9

alisha [4.7K]

M = (y2 - y1) / (x2 - x1)
Call (3,-4) one point
Call (-2,5) the second point.

m = (5- -4) / (-2 - 3)
m = 9/-5 = - 9 / 5

B<<<< answer

7 0

3 years ago

a single carton of juice cost $4.20. A special offer pack of 3 cartons cost $9.45. Jace bought a special offer instead of 3 sing

Ostrovityanka [42]

The 3 pack cost $9.45

3 single cartons would have cost: 3 x 4.20 = $12.60

Difference in cost: 12.60 - 9.45 = $3.15

Percent savings : 3.15/ 9.45 = 0.3333

0.333 x 109 = 33.33%

Round the answer as needed

7 0

2 years ago

The weekly salaries (in dollars) for

kompoz [17]

Answer:

a) Median stays the same

b) Mean is decreased by $9

Step-by-step explanation:

The median is the number or the average of the two numbers that is in the middle of a sorted distribution of numbers,

Here the median number will be the 5th number counting from left or right from the sorted list of numbers. Therefor is is 891.

When 1027 is changed to 946 it will fall between 938 and 1002. So updated sorted list of numbers will now look like,

679, 715, 799, 844, 891, 917, 938, 946, 1002

Here also median will be the 5th number which will be equal to 891.

Therefore, , median will not change.

Mean is the value we get by taking the total value of the salaries and divide it by the number of employees.

In the initial case,

Mean = $\frac{679+715+799+844+891+917+938+1002+1027}{9} =868$

When the salary is changed from $1027 to $946,

Mean= $\frac{679+715+799+844+891+917+938+1002+946}{9} =859$

Therefor we can see that Mean has decreased by $9.

8 0

3 years ago

La potencia que se obtiene de elevar a un mismo exponente un numero racional y su opuesto es la misma verdadero o falso?

malfutka [58]

Answer:

Falso.

Step-by-step explanation:

Sea $d = \frac{a}{b}$ un número racional, donde $a, b \in \mathbb{R}$ y $b \neq 0$ , su opuesto es un número real $c = -\left(\frac{a}{b} \right)$ . En el caso de elevarse a un exponente dado, hay que comprobar cinco casos:

(a) El exponente es cero.

(b) El exponente es un negativo impar.

(c) El exponente es un negativo par.

(d) El exponente es un positivo impar.

(e) El exponente es un positivo par.

(a) El exponente es cero:

Toda potencia elevada a la cero es igual a uno. En consecuencia, $c = d = 1$ . La proposición es verdadera.

(b) El exponente es un negativo impar:

Considérese las siguientes expresiones:

$d' = d^{-n}$ y $c' = c^{-n}$

Al aplicar las definiciones anteriores y las operaciones del Álgebra de los números reales tenemos el siguiente desarrollo:

$d' = \left(\frac{a}{b} \right)^{-n}$ y $c' = \left[-\left(\frac{a}{b} \right)\right]^{-n}$

$d' = \left(\frac{a}{b} \right)^{(-1)\cdot n}$ y $c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{(-1)\cdot n}$

$d' = \left[\left(\frac{a}{b} \right)^{-1}\right]^{n}$ y $c' = \left[(-1)^{-1}\cdot \left(\frac{a}{b} \right)^{-1}\right]^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c = (-1)^{n}\cdot \left(\frac{b}{a} \right)^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left[(-1)\cdot \left(\frac{b}{a} \right)\right]^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left[-\left(\frac{b}{a} \right)\right]^{n}$

Si $n$ es impar, entonces:

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = - \left(\frac{b}{a} \right)^{n}$

Puesto que $d' \neq c'$ , la proposición es falsa.

(c) El exponente es un negativo par.

Si $n$ es par, entonces:

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left(\frac{b}{a} \right)^{n}$

Puesto que $d' = c'$ , la proposición es verdadera.

(d) El exponente es un positivo impar.

Considérese las siguientes expresiones:

$d' = d^{n}$ y $c' = c^{n}$

$d' = \left(\frac{a}{b}\right)^{n}$ y $c' = \left[-\left(\frac{a}{b} \right)\right]^{n}$

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{n}$

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = (-1)^{n}\cdot \left(\frac{a}{b} \right)^{n}$

Si $n$ es impar, entonces:

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = - \left(\frac{a}{b} \right)^{n}$

(e) El exponente es un positivo par.

Considérese las siguientes expresiones:

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = \left(\frac{a}{b} \right)^{n}$

Si $n$ es par, entonces $d' = c'$ y la proposición es verdadera.

Por tanto, se concluye que es falso que toda potencia que se obtiene de elevar a un mismo exponente un número racional y su opuesto es la misma.

3 0

3 years ago