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tatuchka [14]
3 years ago
10

Use algebra to solve the same equation.

Mathematics
1 answer:
Ray Of Light [21]3 years ago
5 0
I don’t get what you mean?
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A balloon rising at a constant rate will rise farther in 60 seconds than in 45 seconds. Which inequality shows this in function
Citrus2011 [14]
Call f(t) the function that gives the rise of the balloon; then the statment translate to:

f(60) > f(45)
8 0
4 years ago
How many zeros does the function f(x) = 2x14 − 14x6 + 27x3 − 13x + 12 have?
s2008m [1.1K]
<span>The function f(x) = 2x^14 − 14x^6 + 27x^3 − 13x + 12 has 2 zeros. </span>
4 0
3 years ago
What is 5 ones , 38 tens , 27 hundreds , and 5 thousands? Write the number in word form and expanded form
weqwewe [10]

<em>Answer:</em>

Eight thousand eighty five, 8000 + 80 + 5

<em>Explanation:
</em>Five ones, thirty-eight tens, twenty-seven hundreds, and five thousands equal to:

5 + 380 + 2700 + 5000

This is 8085 when added or eight thousand eighty-five in word form. This is also 8000 + 80 + 5 in expanded form.

3 0
2 years ago
Un futbolista ha metido 2/9 del número de goles marcados por su equipo y otro la
storchak [24]

Answer:

77 goles

Step-by-step explanation:

Un futbolista ha marcado 2/9 del número de goles marcados por su equipo.

Otro anotó una cuarta parte del resto.

El resto = 1 - 2/9

= 7/9

De ahí que otro futbolista anotó

= 1/4 de 7/9

= 7/36

Si los otros jugadores han marcado 45 goles

Tenemos que averiguar la fracción de goles que marcó el otro jugador

Deje que el número total de goles marcados por el equipo durante la temporada = 1

Por lo tanto:

1 - (2/9 + 7/36)

1 - (8 + 7/36)

1 - 15/36

1 - 5/12

= 7/12

¿Cuántos goles marcó el equipo a lo largo de la temporada?

El número total de goles que marcó ese equipo se calcula como:

7/12 × x = 45

Donde x = número total de goles

7x / 12 = 45

Cruz multiplicar

7x = 45 × 12

x = 45 × 12/7

x = 77,142857143

Aproximadamente = 77 goles

7 0
3 years ago
What is the answer to Factor 3a^2-6a
Anton [14]
Hi there!

3 {a}^{2}  - 6a \:  = 3a(a - 2)

We can see this by writing the factors as multipliers. We get the following:
3a {}^{2}  = 3 \times \: a  \times a
6a = 3 \times 2 \times a

Now we can see that the terms that are in common are 3 and a, and therefore we can bring those outside the parenthesis.

7 0
3 years ago
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