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Contact [7]
3 years ago
7

Tommy saw 25 trains on each of the last 5 days of school. what is the unit rate

Mathematics
2 answers:
zhannawk [14.2K]3 years ago
8 0

Answer:

the unit rate would be once per day he saw 25 trains and each day he adds 25 more trains and on the 5th day he saw 125 trains

Step-by-step explanation:

Ksivusya [100]3 years ago
7 0

Answer:

25 per day

Step-by-step explanation:

it says that everyday he saw 25 so if he saw 25 everyday, then by the 5th day, he had seen a total of 125 trains because 25 per day times 5 days =125 trains in 5 days

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On a science exam, Jackson scored 85 out of 100 possible points. There were 20 questions on the exam. Each question was worth th
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Do 85÷20 and then add it to 100
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3 years ago
If line segment BH = 3, line segment BF = 10, line segment GH= 2, find HD 15 11 10.5 ?
pav-90 [236]

HD = 10.5

Step-by-step explanation:

Given BH = 3, GH = 2, BF = 10

Step 1: To find HF:

HF = BF – BH

HF = 10 – 3

HF = 7

Step 2: To find HD:

We know that if two chords intersects inside a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.

⇒ GH × HD = BH × HF

⇒ 2 × HD = 3 × 7

⇒ HD = 10.5

Hence, the value of HD = 10.5.

5 0
3 years ago
Read 2 more answers
In 2003, there were about 5.8078 X 10^8 people using the Internet in the world and about 1.6575 X 10^8 of these people were in t
lianna [129]

Answer:

28.5%  Internet users in 2003 were in the United States.    

Step-by-step explanation:

We are given the following in the question:

Number of people using the internet =

5.8078 \times 10^8

Number of people in United States =

1.6575 \times 10^8

Percent of Internet users in 2003 were in the United States =

=\dfrac{\text{Number of users in United States}}{\text{Number of internet user in world}}\times 100\%\\\\=\dfrac{ 1.6575 \times 10^8}{5.8078 \times 10^8}\times 100\%\\\\=28.5\%

Thus, 28.5%  Internet users in 2003 were in the United States.

6 0
3 years ago
What is the equation of the quadratic function represented by this table ??HELP PLEASE GRADUATION IS JUNE FIRST
miss Akunina [59]

9514 1404 393

Answer:

  -3, 7, 8

Step-by-step explanation:

You can fill in the boxes from the table values as shown in the attachment.

The leading coefficient is negative because the parabola opens downward. The value of it is the change in y-value as x changes by 1 unit either side of the vertex.

The vertex is identified by the fact that the y-values are symmetrical either side of the vertex.

5 0
3 years ago
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A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th
Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let S(t) be the amount of sugar in the tank at time t. Then S(0)=25.

Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}

Separate variables, integrate, and solve for <em>S</em>.

\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt

\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt

-100\ln\left|P-\dfrac S{100}\right|=t+C

\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t

P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}

\dfrac S{100}=P-Ce^{-100t}

S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}

Use the initial value to solve for <em>C</em> :

S(0)=25\implies 25=100P-C\implies C=100P-25

\implies S(t)=100P-(100P-25)e^{-100t}

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick <em>P</em> such that

S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}

5 0
3 years ago
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