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sveticcg [70]
3 years ago
11

On monday delilahs family drives 45 1/3 miles each hour

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
3 0

Answer:

15 total miles each hour

Step-by-step explanation:

for this situation and the information given in the question, all we have to do is divide the total amount by 3 (since its 1/3) to get the amount of total miles they drove at, wich is the speed they drove in this case

(sidenote: kinda slow if ya ask me but ok XD)

hope this helps you :D

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What is 4 1/4 multiplied by 5 as a mixed number or a fraction
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22 1/2

Step-by-step explanation:


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3 years ago
Which expression is equivalent to one over five m − 20?
Arada [10]

Answer:

1/5m - 20 = 1/5(m - 100)

Step-by-step explanation:

3 0
3 years ago
How much is six dimes, 8 nickels, and three one-dollar bills? *
Alex_Xolod [135]

Answer:

.60 + .40 + 3.00 = 4.00

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Part A: The Sun produces 3.9 ⋅ 1033 ergs of radiant energy per second. How many ergs of radiant energy does the Sun produce in 3
Nuetrik [128]

5914 1404 393

Answer:

  A) 1.3×10^37 ergs

  B) 1.435×10^3 mm

Step-by-step explanation:

A) The amount of energy will be the product of the energy rate and time:

  (3.9×10^33 ergs/s)×(3.25×10^3 s) =12.675×10^(33+3) ergs

  = 1.2675×10^(1+36) ergs

  = 1.2675×10^37 ergs ≈ 1.3×10^37 ergs

The mantissa of the result is the product 3.9×3.25, adjusted to have one digit left of the decimal point. The exponent of the result is the sum of the exponents of the factors, adjusted by 1 to match the adjustment in the mantissa.

The final value should be rounded to 2 significant figures, reflecting the precision of the sun's energy production.

__

B) A millimeter is a small fraction of an inch. 10^-3 mm is a small fraction of the width of a human hair, so 1.435×10^-3 mm is not a reasonable estimate of the distance between railroad tracks.

On the other hand, 1.435×10^3 mm is 1.435 m, almost 56.5 inches. This is a much more reasonable measurement of the distance between railroad rails.

  1.435×10^3 mm is more reasonable

8 0
3 years ago
The area of a rectangular field is 1000 yd2.Two parallel sides are fenced with aluminum at $15/yd. One of the remaining sides is
ioda
Coolio

lw=1000
lets assume that the aluminum sides are the legnth
2 sides
2*15=30
30l=cost of aluminum

steel=10w
wood=(w-10)5=5w-50
we can eliminate the w by solving for w in first relation

lw=1000
divide both sides by l
w=1000/l
sub that for w

10(1000/l)=steel
5(1000/l)-50=wood

wood cost+steel+aluminum cost=total cost
10(1000/l)+5(1000/l)-50+30l=1525
(10000/l)+(5000/l)-50+30l=1525
(15000/l)-50+30l=1525
add 50 to both sides
(15000/l)+30l=1575
times both sides by l
15000+30l²=1575l
minus 1575l from both sides
30l²-1575l+15000=0
factor
(15)(l-40)(2l-25)=0
set each equal to 0
l-40=0
l=40

2l-25=0
2l=25
l=12.5



the there are 2 possible combinations

aluminum=40yd and steel=25yd or
aluminum=12.5yd and steel=80ft
both yield 1000yd² and cost $1525
5 0
4 years ago
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