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ludmilkaskok [199]
3 years ago
13

You have quarters and dimes that total $2.80. your friend says it is possible that the number of quarters is 8 more than the num

ber of dimes. is your friend correct? explain. let dd represent the number of dimes.an equation that models this situation is?
Mathematics
1 answer:
ss7ja [257]3 years ago
4 0
To answer this question, you need to know that a quarter is worth $0.25 and dimes worth $0.10. In this case, the total number of money that made from quarters and dimes is $2.80
With Q=quarter and D=dime, then the equation should be:
2.80= 0.25 Q + 0.10 D

The condition that needs to be fulfilled in this case is that the quarter should be 8 more than dimes. Then you need to find out the highest possible number of the quarter and lowest possible number of the dimes.  
If you divide 2.8 with 0.25 you will found it will be 11.2 but 11 quarter will result with 0.05 value which you cannot remove, so the highest possible of the quarter should be 10.
If the highest possible of the quarter is 10, then the lowest possible of dimes should be:
2.8= 0.25(10) + 0.10D
0.10D= 0.3
D=3

Since 10-3 is 7, then it is not possible for the quarter to be 8 more than the dimes.
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These liness pass through the same point
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Answer:

All of them except B

Step-by-step explanation:

Bisecting is splitting a line in half

Parallel lines never touch each other

Intersecting lines intersect

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8 0
3 years ago
When Ryan counted some dimes and quarters he had 196 coins and a total of $39.40. How many coins of each kind does Ryan have?
Nonamiya [84]
Quarters = x
dimes = 196 - x

10(196 - x) + 25x = 3940
1960 - 10x +25x = 3940
1960 + 15x = 3940
15x = 1980
x = 132

169 - 132 = 64

Therefor there are 132 quarters and 64 dimes
6 0
3 years ago
Calculate the following<br><br>A)-10-(-19)+(8-16)<br><br>B)-6×2+4÷2​
Lelechka [254]

Answer:

A. 1

B. -10

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
You how many miles are there in 86.88 km 1.6 km equals to 1 miles or conversion ​
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86.88 kilometers is 53.98 miles
3 0
3 years ago
UPPER AND LOWER BOUNDS - PLEASE HELP!
Marianna [84]
Qn. 1
Lower bound for Zoe's weight = 62 - (1/2) = 62 - 0.5 = 61.5 kg

Qn. 2
Upper bound for length AB = 8.3+ (0.1/2) = 8.3+0.05 = 8.35 cm

Qn. 3
Upper bound for Anu's wight = 83+(0.5/2) = 83+0.25 = 83.25 kg

Qn. 4
Lower bound for length CD = 27-(0.5/2) = 27-0.25 = 26.75 cm

Qn. 5
Upper bound for sides of the hexagon = 3.6+(0.1/2) = 3.6+0.05 = 3.65 cm
Upper bound for the perimeter = upper bound for the sides*6 = 3.65*6 = 21.9 cm

Qn. 6
Perimeter = 4*length => side = Perimeter/4 = 24/4 = 6
Bound = 0.5/4 = 0.125
Lower bound of the length = 6-0.125 = 5.875 cm

Qn. 7
For the area,
Upper bound = 80+(10/2) 80+5 = 85 cm^2
For the length
Upper bound = 12+(1/2) = 12+0.5 = 12.5

Then, upper bound for the width = Upper bound for the area/upper bound for the length = 85/12.5 = 6.8 cm

Qn. 8
Lower bound for the area = 230-(1/2) = 230-0.5 = 229.5 cm^2
Lower bound for the sides of the square = Sqrt(Lower bound of the area) = Sqrt (229.5) = 15.15
Then,
Lower bound of perimeter = 4(Length) = 4*15.15 = 60.6 cm
8 0
3 years ago
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