QuestionElectricity cost in February was \$93.6, up 50\% from January. Find total cost for both months. What is it?

Studdy Solution
Assumptions1. The electricity cost of February is $93.6. The cost increased by50% from last month (January)
3. We need to find the total cost for these two months (January and February)
Problem11First, we need to find the electricity cost for January. We know that the cost in February is50% more than the cost in January. So, we can write this as$$Cost_{January} = Cost_{February} /1.5$$3Now, plug in the given value for the February cost to calculate the January cost.
$$Cost_{January} = \$93.6 /1.5$$4Calculate the January cost.
$$Cost_{January} = \$93.6 /1.5 = \$62.4$$5Now that we have the cost for January and February, we can add these to find the total cost for the two months.
$$Total\, cost = Cost_{January} + Cost_{February}$$6Plug in the values for the January and February costs to calculate the total cost.
$$Total\, cost = \$62.4 + \$93.6$$7Calculate the total cost.
$$Total\, cost = \$62.4 + \$93.6 = \$156$$The total electricity cost of these two months is $156.
Problem128Assumptions1. The equation given is $10x^{}-x-3=0$ . We need to find the value of $x$
9This is a quadratic equation of the form $ax^{}+bx+c=0$. We can solve it using the quadratic formula$$x = \frac{-b \pm \sqrt{b^{}-4ac}}{a}$$10Now, plug in the values for $a$, $b$, and $c$ from the given equation into the quadratic formula.
$$x = \frac{-(-1) \pm \sqrt{(-1)^{}-4*10*(-3)}}{*10}$$11implify the equation.
$$x = \frac{1 \pm \sqrt{1+120}}{20}$$12Calculate the value of $x$.
$$x = \frac{1 \pm \sqrt{121}}{20}$$$$x = \frac{1 \pm11}{20}$$13So, the solutions are$$x = \frac{1 +11}{20} =0.6$$$$x = \frac{1 -11}{20} = -0.5$$The solutions to the equation $10x^{}-x-3=0$ are $x =0.6$ and $x = -0.5$.