How do you evaluate #z(x+1)# if w=6, x=0.4, y=1/2, z=-3?

2 Answers
May 2, 2017

Answer:

See the solution process below:

Explanation:

To evaluate this expression substitute #color(red)(-3)# for #color(red)(z)# and #color(blue)(0.4)# for #color(blue)(x)#. The values for #w# and #y# are not used in the expression and are extraneous or not needed or necessary:

#color(red)(z)(color(blue)(x) + 1)# becomes:

#color(red)(-3)(color(blue)(0.4) + 1) => color(red)(-3) * 1.4 => -4.2#

May 2, 2017

Answer:

#z(x+1)=color(green)(-4.2)#

Explanation:

Given
#color(white)("XXX")color(orange)w=color(orange)6#
#color(white)("XXX")color(red)x=color(red)(0.4)#
#color(white)("XXX")color(brown)y=color(brown)(1/2)#
#color(white)("XXX")color(blue)z=color(blue)(-3)#

Then
#color(white)("XXX")color(blue)z(color(red)x+1)#
#color(white)("XXXXXXX")=(color(blue)(-3))xx(color(red)(0.4)+1)#
#color(white)("XXXXXXX")=(color(blue)(-3))xx1.4#
#color(white)("XXXXXXX")=-4.2#