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konstantin123 [22]
2 years ago
10

2. A student on the track team runs 45 minutes each day as a part of her training. She

Mathematics
1 answer:
fomenos2 years ago
8 0

Answer:

D

Step-by-step explanation:

The basic equation to solve this would be D = RT, which is

D is Distance

R is Rate (speed)

T is time

To find total distance D, we can find individual distances for two legs of the whole.

First Leg:

R = 8

T = a

D = 8a

Second Leg:

R = 7.5

T = b

D = 7.5b

Total distance D is:

D = 8a + 7.5b

Moreover, we know student runs 45 minutes in total hence a + b = 45 or we can say a = 45 - b, so we can replace this in the equation found:

D = 8a + 7.5b

D = 8(45 - b) + 7.5b

Answer choice D is right.

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The function D(t) defines a traveler's distance from home, in miles, as a function of time, in hours. D(t) = StartLayout enlarge
nalin [4]

- At 2 hours, the traveler is 725 miles from home.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE.

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE.

Step-by-step explanation:

The function D(t) is defined as follows:

D(t) = 300t+125 for t< 2.5

D(t) = 880 for 2.5 \leq 3.5

D(t) = 75t+612.5 for t\leq 6

Where

t is the time in hours

D(t) is the distance covered, in miles, after t hours

Now let's analyze the different statements:

- The starting distance, at 0 hours, is 300 miles. --> FALSE. In fact, if we substitute t = 0 into the 1st equation, we get

D(0) = (300)(0)+125 = 125

So, the distance at t = 0 is 125 miles.

- At 2 hours, the traveler is 725 miles from home. --> TRUE. In fact, if we substitute t = 2 into the 1st equation,

D(2) = (300)(2)+125 = 725

- At 2.5 hours, the traveler is 875 miles from home. --> FALSE. In fact, for t=2.5 we have to use the 2nd equation, which states that the distance is:

D(t) = 880

So, not 875 miles.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE. This is clearly visible from the 2nd equation: for t between 2.5 and 3.5 (so, in this case), the distance is

D(t) = 880

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE. In fact, if we replace t = 6 into the last equation,

D(6)) = 75(6)+612.5=1062.5

Learn more about functions:

brainly.com/question/3511750

brainly.com/question/8243712

brainly.com/question/8307968

Learn more about  distance:

brainly.com/question/3969582

#LearnwithBrainly

4 0
1 year ago
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Six Flags is offering a package deal of 5 entrance passes for $130. If 1 entrance pass normally costs $30. how much will you sav
stiv31 [10]

Answer:

You will save $20

Step-by-step explanation:

30x5=150-130=$20

7 0
2 years ago
21.05 divided by 0.2
Valentin [98]

Answer:

Step-by-step explanation:

8 0
1 year ago
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What additional information could be used to prove that ΔXYZ ≅ ΔFEG using ASA or AAS? Check all that apply.
Maksim231197 [3]

Answer: ∠Z ≅ ∠G and XZ ≅ FG or ∠Z ≅ ∠G and XY ≅ FE  are the additional information could be used to prove that ΔXYZ ≅ ΔFEG using ASA or AAS.

Step-by-step explanation:

Given: ΔXYZ and ΔEFG such that ∠X=∠F

To prove they are congruent by using ASA or AAS conruency criteria

we need only one angle and side.

1. ∠Z ≅ ∠G(angle) and XZ ≅ FG(side)

so we can apply ASA  such that ΔXYZ ≅ ΔFEG.

2. ∠Z ≅ ∠G (angle)and ∠Y ≅ ∠E (angle), we need one side which is not present here.∴we can not apply ASA  such that ΔXYZ ≅ ΔFEG.

3. XZ ≅ FG (side) and ZY ≅ GE (side), we need one angle which is not present here.∴we can not apply ASA  such that ΔXYZ ≅ ΔFEG.

4. XY ≅ EF(side) and ZY ≅ FG(side), not possible.

5. ∠Z ≅ ∠G(angle) and XY ≅ FE(side),so we can apply ASA  such that

ΔXYZ ≅ ΔFEG.

4 0
2 years ago
Read 2 more answers
The following table shows the number of hours some teachers in two schools expect students to spend on homework each week: Schoo
Lerok [7]

Answer:

For school A: Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17, IQR=9.5

For school B: Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19, IQR=7.5

No, the box plots are not symmetric.

Step-by-step explanation:

Part A

The given data sets are

School A : 9,14,15,17,17,7,15,6,6

School B : 12,8,13,11,19,15,16,5,8

Arrange the data in ascending order.

School A : 6,6,7,9,14,15,15,17,17

School B : 5,8,8,11,12,13,15,16,19

Divide each data set in four equal parts.

School A : (6,6),(7,9),14,(15,15),(17,17)

School B : (5,8),(8,11),12,(13,15),(16,19)

For school A:

Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17

Interquartile range of the data is

IQR=Q_3-Q_1=16-6.5=9.5

For school B:

Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19

Interquartile range of the data is

IQR=Q_3-Q_1=15.5-8=7.5

Part B:

The box plots are not symmetric because the data values are different. Five number summary and IQR of both the data set are different.

4 0
2 years ago
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