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MA_775_DIABLO [31]
3 years ago
6

Can someone that will GIVE ME A GOOD ANSWER can you help me please

Mathematics
2 answers:
Natali5045456 [20]3 years ago
8 0

Answer: -3 for the exponent box

Step-by-step explanation: Hi there! Let's get started.

We know that a negative exponent is going to decrease by a power of 10 every time for example 10 x 10^-1 = 1 because the -1 says we need to move the decimal point one to the left.

So with this we take 2.345 x 10^-3 = 0.002345.

We moved the decimal point 3 to the left from 2.345 to get 0.002345.

Hope this helps! :)

jenyasd209 [6]3 years ago
4 0

Answer:  -3 bc the exponent is adding -1 each time

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En un acuario compraron 30 peces y 18 tortugas de agua. Los van a colocar en peceras. Cada una tiene que haber la misma cantidad
seraphim [82]

Answer:

a) ¿Cuántas peceras compraron?

Según el cálculo anterior, la cantidad de peceras que compró es de 90 peceras.

b) ¿Cuántos peces y tortugas puede tener cada tanque?

3 peces por tanque

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Step-by-step explanation:

Resolvemos la pregunta anterior utilizando el cálculo de múltiplo común más bajo

En un acuario compraron 30 peces y 18 tortugas de agua.

Encuentre y enumere los múltiplos de cada número hasta encontrar el primer múltiplo común. Este es el mínimo común múltiplo.

El mínimo común múltiplo es el número de tanques

Múltiplos de 18:

18, 36, 54, 72, 90, 108, 126

Múltiplos de 30:

30, 60, 90, 120, 150

Por lo tanto,

MCM (18; 30) = 90

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Según el cálculo anterior, la cantidad de peceras que compró es de 90 peceras.

b) ¿Cuántos peces y tortugas puede tener cada tanque?

El número de peces que puede tener cada tanque = Número de tanques / Número de peces

= 90/30

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El número de tortugas de agua que puede tener cada tanque =

Número de tanques / Número de tortugas de agua

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5 0
3 years ago
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zimovet [89]
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If the discriminant is greater than 0, the quadratic equation has 2 real different roots
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If the discriminant is less than 0, the quadratic equation doesn't have real root.

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D = 0
The equation has only one real root.
5 0
4 years ago
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