Answer: The value of the card after you buy your 8th coffee will be $61.3
Step-by-step explanation:
The worth of the gift card for the coffee shop is $90. Each day you use the card to get a coffee for $4.10. This means that the worth of the gift card is reducing by $4.10 each day. This rate is in arithmetic progression.
The formula for the nth term of an arithmetic sequence, Tn is expressed as
Tn = a + (n-1)d
Where a is the first term
d is the common difference
n is the number of days
From the information given,
a = $90
d = - $4.1
The explicit formula representing the amount of money available will be
Tn = 90 - 4.1(n - 1)
The value of the card after you buy your 8th coffee will be
T8 = 90 - 4.1(8 - 1) = T8 = 90 - 4.1×7
T8 = 90 - 28.7
T8 = $61.3
Answer: 12.85
Step-by-step explanation: If you start off by multiplying them then. You'd get 225=20x-2. So from there you subtract the 2, to isolate the 20x. Then divide bother sides by 20 to get 12.85.
Answer:
let the breadth of room be 'x'm
so the lenghth of room is 'x+8'm
Given
Area of room=48m²
i.e length ×breadth =48m²
or,(x+8)x=48
or,x²+8x-48=0
or,x²+12x-4x-48=0
or,x(x+12)-4(x+12)=0
or,(x+12)(x-4)=0
Either x+12=0
or,x=-12(which is not possible)
Or,x-4=0
or,x=4
Thus,breadth is 4m.
length=x+8=4+8=12m
The dimensions of room are 8m and 12m respectively.
7 miles:2.5 hours
10 miles:? hour
10×2.5÷7=25/7. As a result, It took him 25/7 hours or 3 4/7 hours to ride 10 miles. Hope it help!
9514 1404 393
Answer:
the difference of 2 or 3 rectangles
Step-by-step explanation:
In every case, the "shaded" area can be computed by finding the area of a "bounding" rectangle, and subtracting the areas of the rectangular cutouts that give the figure its shape.
(a) The cutout is the white space at upper right. (Insufficient dimensions are given.)
(b) The cutout is the white space at lower left. The bounding rectangle is 8×7, and the cutout is 4×3.
(c) The cutout is the rectangle in the middle. The bounding rectangle is 13×7, and the cutout is 4×1.
(d) The cutouts are the rectangles on either side. They could be considered as a single unit. The bounding rectangle is 20×25; the cutouts have a total width of 16 and a height of 20, so total 16×20.
(e) Similar to (d), the cutouts are the white spaces on either side. The bounding rectangle is 14×12. The cutouts total 12 in width and 3 in height, so total 12×3.
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You will note in (d) and (e) that the dimensions of the cutouts have something in common with the dimensions of the bounding rectangle. This means the problem can be simplified a little bit by factoring out that common factor. In (e), for example, 14×12 -(12×3) = 12(14 -3) = 12×11
