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SVETLANKA909090 [29]
3 years ago
6

Judy is playing a card game. She needs to select two cards below that have a sum greater

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
7 0

Answer:

2\frac{2}{5}  + 2\frac{5}{8} \simeq 5

2\frac{2}{7}  + 2\frac{5}{8}  \simeq 5

Step-by-step explanation:

Mixed fraction a\frac{b}{c} can be converted into like or unlike fractions.

Here, a\frac{b}{c} =  \frac{ac +b}{b}

So, convert all the given fractions as follows:

2\frac{2}{5} =  \frac{2(5) +2}{5} = \frac{12}{5}  =2.4

2\frac{2}{7} =  \frac{2(7) +2}{7} = \frac{16}{7}  =2.4

1\frac{7}{8} =  \frac{1(8) +7}{8} = \frac{15}{8}  =1.8

2\frac{1}{10} =  \frac{2(10) + 1}{10} = \frac{21}{10}  =2.1

2\frac{5}{8} =  \frac{2(8) +5}{8} = \frac{21}{8}  =2.6

Here, No Two fractions on addition gives a sum greater than 5.

But an approximate sum of 5 is given by adding 2.4 + 2.6 = 5

⇒2\frac{2}{5}  + 2\frac{5}{8} \simeq 5

or  ⇒2\frac{2}{7}  + 2\frac{5}{8}  \simeq 5

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Answer:

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Step-by-step explanation:

8 0
3 years ago
Solve the equation of exponential decay.
BabaBlast [244]

Answer:

$9,220,000(0.888)^t

Step-by-step explanation:

Model this using the following formula:

Value = (Present Value)*(1 - rate of decay)^(number of years)

Here, Value after t years = $9,220,000(1 -0.112)^t

          Value after t years =  $9,220,000(0.888)^t

3 0
3 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
Simplify square root of five times the quantity six minus four square root of three.
NemiM [27]

6√5 - 4√3-

i took the test but sorry if its wrong

5 0
3 years ago
Steven has some money. If he spends $9.00, then he will have 3 5of the amount he started with
docker41 [41]
$1.80 if the question was ( he would have 3/5 of the amount he started with)
7 0
2 years ago
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