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KATRIN_1 [288]
2 years ago
11

Please help, im trying get my grade up to go on a date w my bf

Mathematics
2 answers:
Inga [223]2 years ago
7 0

Answer:

a+4b-24

Step-by-step explanation:

You have to multiply the entire equation by 4, which get you a+4b-24

const2013 [10]2 years ago
3 0

Answer:

a+4b-24

Step-by-step explanation linked:

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please help i don't understand this one i appreciate anyone who helps :) Solve for x. 7x - 3/8 = 6x -
svet-max [94.6K]

Answer:

x = 3/8

Step-by-step explanation:

6 0
3 years ago
What is the length of the unknown side?
777dan777 [17]

Answer:

Hi there!

The correct answer is: 20

Step-by-step explanation:

knowing this a right triangle you can solve this problem in two ways

Method One: Pythagorean Theorem

a^2 + b^2 = c^2 then plug in the values

(12)^2 + (16)^2 = c^2 this will come out to be 400 = c^2

square root both sides and you get c = 20

Method Two: Pythagorean Identities

if you ever learned the Pythagorean identity 3,4,5

this triangle is indeed a 3,4,5 triangle it's just that each side is multiplied by the factor 4

so in this case since you know the missing side should be 5 you just multiply 5 by 4 and you get 20

 

4 0
3 years ago
In which number is the digit 5 one hundred times larger than it is in the number 253?
Paraphin [41]
The answer is B.
I timesed 50 by 100 and 1,000. So 5 would be in the thousands place.
7 0
2 years ago
Read 2 more answers
What is 32 divided by 6.79
TiliK225 [7]

Answer:

32 divided by 6.79 is 4.71281296024

Step-by-step explanation:

Brainliest please?

5 0
3 years ago
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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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