Carlos has \dfrac45 5 4 start fraction, 4, divided by, 5, end fraction of a tank of fuel in his car. he uses \dfrac{1}{10} 10 1 start fraction, 1, divided by, 10, end fraction of a tank per day. how many days will his fuel last?

  8 days

Step-by-step explanation:

You can rewrite the original fraction with a denominator of 10 to see the answer.

  4/5 = 8/10 = 8·(1/10)

The fuel will last 8 days.

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

solve systems of equations by graphing. a system of linear equations contains two or more equations e.g. y=0.5x+2 and y=x-2. the solution of such a system is the ordered pair that is a solution to both equations. to solve a system of linear equations graphically we graph both equations in the same coordinate system