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kakasveta [241]
3 years ago
6

Ruben has decided that all even number are composite since they can all be dovided by 2.Is Ruben correct?explain

Mathematics
1 answer:
IRISSAK [1]3 years ago
5 0
Yes. All even numbers are divisible by 2. Every single one of them
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One pound of grapes costs $1.55. Which equation correctly shows a pair of equivalent ratios that can be used to find the cost of
stira [4]

Answer:

StartFraction 1.55 over 1 EndFraction = StartFraction x over 3.5 EndFraction

Step-by-step explanation:

\frac{1.55}{1}  =  \frac{x}{3.5}  \\  \\  \therefore \: x = 1.55 \times 3.5 \\  \\ \therefore \: x = \$ 5.425 \\  \\\therefore \: x\approx  \$5.43

4 0
2 years ago
Read 2 more answers
X2+4x+3 can someone explain ?
Katyanochek1 [597]

Answer:

Factor: (x+1) (x+3)

Step-by-step explanation:

8 0
3 years ago
The point A(-3,5) and the point B(1,-15) lie on the line L. Find the equation of the line L.
Shkiper50 [21]
(-3,5)(1,-15)
slope = (-15 - 5) / (1 - (-3) = -20/4 = -5

y = mx + b
slope(m) = -5
use either of ur points...(-3,5)...x = -3 and y = 5
now we sub and find b, the y int
5 = -5(-3) + b
5 = 15 + b
5 - 15 = b
-10 = b
so ur equation is : y = -5x - 10....or 5x + y = -10
7 0
3 years ago
Shirts are on sale for 20% you buy two of them. When you pay the Cheshire says ""20% off each shirt means 40% off the total pric
MAXImum [283]

Answer:

True

Step-by-step explanation:

2 shirts where both at 20% off for each =

          2 * 20% = 40% i.e. 40% off for both shirts  

5 0
3 years ago
Escribe la posición del móvil si el diámetro de la trayectoria es de 10 m y la distancia recorrida es de 190 m
lions [1.4K]

Step-by-step explanation:

La posici´on de una part´ıcula que se mueve unidimensionalmente esta definida por la ecuaci´on:

x(t) = 2t

3 − 15t

2 + 24t + 4 donde 0x

0 y

0

t

0

se expresan en metros y segundos respectivamente. Determine:

a. ¿Cu´ando la velocidad es cero?

b. La posici´on y la distancia total recorrida cuando la aceleraci´on es cero.

Soluci´on:

a. Recordemos que:

v(t) =

dx

dt =

d

dt(2t

3 − 15t

2 + 24t + 4) = 6t

2 − 30t + 24

Sea t

0

el tiempo en que la velocidad se anula, entonces v(t

0

) = 0.

De este modo:

0 = v(t

0

) = 6(t

0

)

2 − 30(t

0

) + 24 = 6[(t

0

)

2 − 5(t

0

) + 4] = 6[(t

0

) − 4][(t

0

) − 1]

As´ı tenemos que:

t

0

1 = 4, t

0

2 = 1

De este modo, tenemos que la velocidad se anula al primer segundo y a los cuatro segundos.

b. Recordemos que:

a(t) =

dv

dt =

d

dt(6t

2 − 30t + 24) = 12t − 30

Ahora sea t

0

el instante en que la aceleraci´on se anula, entonces a(t

0

) = 0

Ahora:

0 = a(t

0

) = 12t

0 − 30

As´ı tenemos que: t

0 =

30

12 =

5

2

Por lo tanto, la posici´on en este instante es:

x(t

0

) = x

5

2

= 2

5

2

3 − 15

5

2

2 + 24

5

2

+ 4 = 125

4 − 3

125

4 + 60 + 4 = −2

125

4 + 64 = −

125

2 +

128

2 =

3

2

De este modo, la posici´on de la part´ıcula cuando la aceleraci´on es cero es de 3

2 metros.

Adem´as la distancia total recorrida esta dada por:

distancia = |x(t

0

) − x(0)| = |

3

2 − 4| =

5

2

Finalmente la distancia total recorrida es: 5

5 0
2 years ago
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