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GaryK [48]
3 years ago
15

To arrive to his appointment on time, Mr. Jones had to drive all the way from his home with the average speed of 60 mph. Due to

heavy traffic, he was driving 15 mph slower than he planned and arrived to the appointment 20 minutes later. How many miles from Mr. Jones' home was his appointment?
Mathematics
1 answer:
FinnZ [79.3K]3 years ago
8 0
35,35 35353553535353553535353535535353535535353553
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Como convertir la oración en una ecuación algebraica
zimovet [89]

Ok, entonces creo que sería -6 porque -6 x 3 = -16, no estoy seguro de si esto ayudará o te confundirá más, pero ten un buen día

6 0
3 years ago
The predicted cost c (in hundreds of thousands of dollars) for a company to remove p% of a chemical from its waste water is show
alukav5142 [94]
We presume your cost function is
  c(p) = 124p/((10 +p)(100 -p))

This can be rewritten as
  c(p) = (124/11)*(10/(100 -p) -1/(10 +p))

The average value of this function over the interval [50, 55] is given by the integral
  \frac{1}{55-50} \times \frac{-124}{11} \int\limits^{55}_{50} {(\frac{1}{x+10}+\frac{10}{x-100})} \, dx
This evaluates to
  (-124/55)*(ln(65/60)+10ln(45/50)) ≈ 2.19494

The average cost of removal of 50-55% of pollutants is about
  $2.19 hundred thousand = $219,000

8 0
3 years ago
??????????????????????????????
Anna007 [38]

You want to find the time <em>t</em> such that

\$1000 = \$200 e^{0.04t}

Divide both sides by $200 :

\dfrac{\$1000}{\$200} = 5 = e^{0.04t}

Take the logarithm of both sides:

\ln(5) = \ln(e^{0.04t}) = 0.04t\ln(e) = 0.04t

Divide both sides by 0.04 and solve for <em>t</em> :

t = \dfrac{\ln(5)}{0.04} \approx 40.24 \approx \boxed{40}

6 0
3 years ago
Unit Rate ez pls help
aliya0001 [1]

Answer:

it can print 24 pages in 15 minutes which means it can print 72 pages in an 3/4 of an hour. 96 in 1 hour, and

Step-by-step explanation:

3 0
3 years ago
Haley walked a total of 6 kilometers by making 2 trips to school. How many trips will Haley
icang [17]

Answer: 6 trips

Step-by-step explanation:

8 0
3 years ago
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