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2 years ago

Assuming you have a TCF Free Student Checking Account, how many maximum overdraft fees can you incur in one day?

Mathematics

1 answer:

Mkey [24]2 years ago

3 0

Answer:

Overdraft Fee:

We will limit to 7 the combined number of these Overdraft Fees and/or Returned Item Fees charged on any one day, up to a total of $259. If your account remains overdrawn for 4 consecutive business days, on the 5th business day we will charge a Sustained Overdraft Fee of $30

Step-by-step explanation:

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Match each pair of points to the equation of the line that is parallel to the line passing through the points.

Paraphin [41]

we know that

If two lines are parallel, then, their slopes are equal.

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we will proceed to calculate the slope in each case, to determine the solution of the problem

Case A) Point $B(5,2)\ C(7,-5)$

Find the slope BC

Substitute the values in the formula

$m=\frac{-5-2}{7-5}$

$m=\frac{-7}{2}$

$m=-3.5$

The equation $y=-3.5x-15$ is parallel to the line passing through the points $B(5,2)\ C(7,-5)$

therefore

the answer Part A) is

$B(5,2)\ C(7,-5)$ ------> $y=-3.5x-15$

Case B) Point $D(11,6)\ E(5,9)$

Find the slope DE

Substitute the values in the formula

$m=\frac{9-6}{5-11}$

$m=\frac{3}{-6}$

$m=-0.5$

The equation $y=-0.5x-3$ is parallel to the line passing through the points $D(11,6)\ E(5,9)$

therefore

the answer Part B) is

$D(11,6)\ E(5,9)$ ------> $y=-0.5x-3$

Case C) Point $F(-7,12)\ G(3,-8)$

Find the slope FG

Substitute the values in the formula

$m=\frac{-8-12}{3+7}$

$m=\frac{-20}{10}$

$m=-2$

Any linear equation with slope $m=-2$ will be parallel to the line passing through the points $F(-7,12)\ G(3,-8)$

Case D) Point $H(4,4)\ I(8,9)$

Find the slope HI

Substitute the values in the formula

$m=\frac{9-4}{8-4}$

$m=\frac{5}{4}$

$m=1.25$

The equation $y=1.25x+4$ is parallel to the line passing through the points $H(4,4)\ I(8,9)$

therefore

the answer Part D) is

$H(4,4)\ I(8,9)$ ------> $y=1.25x+4$

Case E) Point $J(7,2)\ K(-9,8)$

Find the slope JK

Substitute the values in the formula

$m=\frac{8-2}{-9-7}$

$m=\frac{6}{-16}$

$m=-0.375$

Any linear equation with slope $m=-0.375$ will be parallel to the line passing through the points $J(7,2)\ K(-9,8)$

Case F) Point $L(5,-7)\ M(4,-12)$

Find the slope LM

Substitute the values in the formula

$m=\frac{-12+7}{4-5}$

$m=\frac{-5}{-1}$

$m=5$

The equation $y=5x+19$ is parallel to the line passing through the points $L(5,-7)\ M(4,-12)$

therefore

the answer Part F) is

$L(5,-7)\ M(4,-12)$ ------> $y=5x+19$

8 0

2 years ago

Read 2 more answers

Erika and Rita have added a 1-mile walk to their daily exercise schedule. The table lists the time each of them took to walk the

katovenus [111]

There are two sets of data: Erika's and Rita's. Now, we have to find the average values for each of both and compare them. I would assume that the order of the presentation of data points is alternating, first for Erika followed by Rita. To illustrate what I mean, let's say in Day 1, Erika's time is 20 and Rita's time is 21.25. In Day 2, Erika's time is 21.5 and Rita's is 24.75. And so on and so forth. The complete tabulation is shown in the picture.

Erika's average = (20+21.5+22.75+23.25+20.25)/5 = 21.55
Rita's average = (21.25+24.75+23+23+20.75)/5 = 22.55

On average, Erika is much slower than Rita by 1 minute.

7 0

2 years ago

A 12-centimeter rod is held between a flashlight and a wall as shown. Find the length of the shadow on

choli [55]

Answer:

48 cm

Step-by-step explanation:

Given:

Distance of rod from the wall = 45 cm

Distance of rod from the light = 15 cm

Length of rod = 12 cm

We can see that <DAM and <BAF are equal

Also, <DMA and <BFM are equal because they are corresponding angles

To find the length of the shadow, let's take the equation

$\frac{DM}{BF} = \frac{AM}{A.F}$

Where.:

DM = ½ of length of the rod = ½*12 = 6

A.F = 15 + 45 = 60 cm

AM = 15 cm

Therefore,

$\frac{DM}{BF} = \frac{AM}{A.F}$

$= \frac{6}{BF} = \frac{15}{60}$

Cross multiplying, we have:

15 * B.F = 60 * 6

15 * B.F = 360

$BF = \frac{360}{15}$

BF = 24 cm

The shadow on the wall =

2 * BF

= 2 * 24

= 48 cm

The shadow on the wall is 48 cm

7 0

2 years ago

the area of a concrete rectangular pathway is 350m square and its perimeter pathway is 90 m.what is the length of the pathway?

statuscvo [17]

X, y - the lengths of the sides of the pathway

The area of a rectangle is the product of the lengths of its sides. The area is 350 m².
$xy=350$

The perimeter of a rectangle is two times the sum of the lengths of its sides. The perimeter is 90 m.
$2(x+y)=90 \ \ \ \ \ \ |\div 2 \\
x+y=45 \ \ \ \ \ \ \ \ \ \ |-x \\
y=45-x$

Substitute 45-x for y in the first equation:
$x(45-x)=350 \\
45x-x^2=350 \\
-x^2+45x-350=0 \\
-x^2+10x+35x-350=0 \\
-x(x-10)+35(x-10)=0 \\
(-x+35)(x-10)=0 \\
-x+35=0 \ \lor \ x-10=0 \\
x=35 \ \lor \ x=10$

$y=45-x \\
\hbox{for } x=35: \\
y=45-35=10 \\ \\
\hbox{for } x=10: \\
y=45-10=35$

The dimensions of the pathway are 10 m by 35 m.
The length of the pathway is 35 m.

5 0

2 years ago

For the upcoming holiday season, Dorothy wants to mold 20 bars of chocolate into tiny pyramids. Each bar of chocolate contains 6

vlada-n [284]

Answer:

The largest number of pyramids she can make from 20 bars of chocolate is:

A. 180

Step-by-step explanation:

Firstly, you must find the volume of the 20 bars of chocolate, remember that each bar contains 6 cubic inches of chocolate (6 in^3), so:

Volume of the bars = 6 in^3 * 20
Volume of the bars = 120 in^3

Now, you must find the volume of each pyramid, using the next formula:

Volume of a pyramid = 1/3 B * h

Where:

B = area of the square base.

h = height of the pyramid.

How these values are given in the problem, we only replace the values in the formula:

Area of a pyramid = 1/3 (1 in^2) * 2 in
Area of a pyramid = 2/3 in^3

Finally, we divide the volume of the bars in the volume of each pyramid:

Total number of pyramids = 120 in^3 / 2/3 in^3
Total number of pyramids = 180

5 0

2 years ago