What expression represents the profit, and what is the profit if 240 cell phones are sold?

when jane uses her calling card overseas, the cost of a phone call is .75 for the first three minutes and. 12 for each additiona

artcher [175]

26 minutes, would = 3.51. if you add .12 to that it is 3.63.
so the most that she could talk would be 26 minutes.

7 0

3 years ago

If parallelogram RSTV has diagonals then RSTV must also be a _______. square and a rectangle rhombus and a rectangle rhombus rec

andrew11 [14]

<span>a rectangle rhombus(parallelogram) since even a parallelogram has diagonals</span>

5 0

3 years ago

Graph the inequality: –10 < t < 5

shusha [124]

Answer:

no lo sé perdón buen día puto hgfhkghhhddj

Step-by-step explanation:

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ocvjñxhlkzxhllzhzhlcbvg,igxpgofypyfñjguñcñjgjñfodoyfpugpifojvvvbbbbbbbbbbbbbbbnnputo

7 0

3 years ago

Let f(x) = -2x + 4 and g(x) = 5x - 7. Find f(x) - g(x).

balandron [24]

Answer:

If you are just looking for a simplified answer then the answer is -7x - 3.

4 0

3 years ago

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,

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

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

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

8 0

3 years ago