Mae has $189 in her account. A $7 fee is charged each month the balance is below $750.She withdraw $315. If she makes no deposits or withdraws for the next x months express her balance algebraically

The balance of Mae expressed algebraically will be 189 - 315 - 7x.

The following can be deduced from the question:

Therefore, the amount that is left in her account balance will be:

= 189 - 315 - (7 × x)

= 189 - 315 - 7x

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$189 - $315 - $7x

Step-by-step explanation:

Given the following :

Amount in Mae's account = $189

Each month balance is below $750;

Amount charged = $7

She withdrew $315 from her account and makes no deposit or withdrawal for the next x months

Her balance :

Initial amount = $189

Withdrawal made = $315

Amount charged when balance falls below $750 = fee * number of months = 7 * x = 7x

Hence, her balance after x months will be :

$189 - $315 - $7x

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step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."