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Andre's house, the middle school, and the library are along the same street. The middle school
is between Andre's house and the library. The distance between Andre's house and the middle
school is x + 3, and the distance between the middle school and the library is 4x + 12.
Use the following diagram to write and simplify an expression for the distance between
Andre's house and the library.

PLS HELP!!!!!!​

Respuesta :

calculista

Answer:

The expression for the distance between  Andre's house and the library is

[tex](5x+15)[/tex]

Step-by-step explanation:

Let

A -----> Andre's house

B ----> Middle school

C ----> Library

see the attached figure to better understand the problem

we know that

[tex]AC=AB+BC[/tex] -----> by addition segment postulate

we have

[tex]AB=(x+3)\ units[/tex] -----> distance between Andre's house and the middle  school

[tex]BC=(4x+12)\ units[/tex] -----> distance between the middle school and the library

substitute

[tex]AC=(x+3)+(4x+12)[/tex]

Combine like terms

[tex]AC=(5x+15)\ units[/tex]

therefore

The expression for the distance between  Andre's house and the library is

[tex](5x+15)[/tex]

Ver imagen calculista

Answer:

x+3+4x+12

Step-by-step explanation:

WIK:

. Andres house, the middle school and the library are along the same street.

. The middle school is between Andre's house and the library.

.The distance between Andre's house and the middle

school is x+3

. The distance between the middle school and the library is 4x+12

That means that you have to add x+3 +  4x + 12, since the distance between  the middle school is between Andre's house and the library is x+3, and the distance between  the middle school and the library is 4x+12. That means we have to add since the middle school and the library are along the same street.

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