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Marat540 [252]
4 years ago
11

Texas chic this is for you also

Mathematics
1 answer:
Marina86 [1]4 years ago
6 0
The mean because the data is symmetrical and there are no outliers
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1. What is the expanded form of the expression 3.6(t + 5)?
Fed [463]

Answer:

Step-by-step explanation:

1) The expanded form of 3.6(t + 5) is 3.6t + 18.

7 0
3 years ago
Read 2 more answers
Sistema de ecuaciones.5x+2y=-152x-2y=-6
Zolol [24]

Tnemos el sisema de ecuaciones:

\begin{gathered} 5x+2y=-15 \\ 2x-2y=-6 \end{gathered}

Podemos resolverlo por eliminación sumando ambas ecuaciones y eliminando y. Asi podemos resolver para x:

\begin{gathered} (5x+2y)+(2x-2y)=(-15)+(-6) \\ 7x+0y=-21 \\ x=-\frac{21}{7} \\ x=-3 \end{gathered}

Ahora podemos resolver para y con cualquiera de las dos ecuaciones:

\begin{gathered} 2x-2y=-6 \\ 2\cdot(-3)-2y=-6 \\ -6-2y=-6 \\ -2y=-6+6 \\ -2y=0 \\ y=0 \end{gathered}

Respuesta: x=-3, y=0

7 0
1 year ago
Select the correct answer from the each drop-down menu. if 0
diamong [38]

Answer:

You didn't provide a image of the problem how am I suppose to help you

8 0
3 years ago
Which other congruency statements are true?
horrorfan [7]

ans 4th. ∆dcj =∆yam

8 0
3 years ago
what is the quotient 5-x/x^2 3x-4 divided by x^2-2x-15/x^2 5x 4 in simplifed form state any restrictions on the varible
zheka24 [161]

The quotient when \frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4} in simplified form is \frac{-(x+1)}{(x-1)(x+3)}

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that equation:

\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}

=\frac{5-x}{(x+4)(x-1)} /\frac{(x-5)(x+3)}{(x+4)(x+1)} \\\\=\frac{5-x}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\\frac{-(x-5)}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\=\frac{-(x+1)}{(x-1)(x+3)}

The quotient when \frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4} in simplified form is \frac{-(x+1)}{(x-1)(x+3)}

Find out more on equation at: brainly.com/question/2972832

#SPJ1

8 0
3 years ago
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