What is the value of the algebraic expression if x= 1/2 , y = -1, and z = 2? 6x(y2z) A. -12 B. -6 C. 1 D. 6

1 answer:

4 0

To determine the value of the given algebraic expression above, we simply substitute the values of each variable to the variables in the expression and evaluate the expression. We do as follows:

6x(y^2)(z)
when x = 0.5
y = -1
z = 2
6(0.5)((-1)^2)(2)
3(1)(2)
6

The value of the algebraic expression would be 6. An algebraic expression consists of variables which are represented with letters like for this case x, y and z, a coefficient which is indicated by numbers (e.g. 6 ) and exponents like 2 for the expression above. Often times expressions contains a number of terms which consists of those elements.

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bx%2B1%7D" id="TexFormula1" title="\frac{x}{x+1}" alt="\frac{x}{x+1}" align="a

navik [9.2K]

Answer:

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What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$