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Masja [62]
2 years ago
5

Ainsley lives in a town 360 miles directly north of Orlando, and one Saturday, she takes the train from her town to the city. he

train travels at a constant speed, and after 2.5 hours, she sees a sign that states "Orlando: 210 miles". What is an equation that represents (x), the distance Ainsley is from Orlando after x hours.
Mathematics
1 answer:
iragen [17]2 years ago
5 0
By definition we have to:
 d = v * t
 where,
 d: distance
 v: speed
 t: time
 We have then that the speed is:
 (360-210) = v * 2.5
 Clearing we have:
 v = (360-210) /2.5
 v = 60 mph
 We now write the generic equation of distance:
 d (x) = 60x
 Where,
 x: time
 Answer:
 
an equation that represents (x), the distance Ainsley is from Orlando after x hours is:
 
d (x) = 60x
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Harriet earns the same amount of money each day. Her gross pay at the end of 7 workdays is 35h+56 dollars.which expression repre
statuscvo [17]

The expression 5h+8 represents Harriet's gross pay each day.

Step-by-step explanation:

Given,

Gross pay of Harriet = 35h + 56

This expression represent Harriet's pay for 7 day.

Harriet's gross pay per day = \frac{Gross\ pay}{Number\ of\ days}

Harriet's gross pay per day = \frac{35h+56}{7}

Harriet's gross pay per day = \frac{35h}{7}+\frac{56}{7}

Harriet's gross pay per day = 5h+8

The expression 5h+8 represents Harriet's gross pay each day.

Keywords: division, addition

Learn more about division at:

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4 0
2 years ago
The first terms of an infinite geometric sequence, Un are 2, 6, 18, 54... The first terms of a second infinite geometric sequenc
gladu [14]

Answer:

r = 9 and m = 112

Step-by-step explanation:

\sum_{k=1}^{225}W_{k}=\sum_{k=0}^{m}4r^{k}

Write W in terms of U and V.

\sum_{k=1}^{225}(U_{k}+V_{k})=\sum_{k=0}^{m}4r^{k}\\\sum_{k=1}^{225}U_{k}+\sum_{k=1}^{225}V_{k}=\sum_{k=0}^{m}4r^{k}

Define U and V using geometric series formula.

\sum_{k=1}^{225}2(3)^{k-1}+\sum_{k=1}^{225}2(-3)^{k-1}=\sum_{k=0}^{m}4r^{k}

Use sum of geometric series formula.

2(\frac{1-(3)^{225}}{1-3})+2(\frac{1-(-3)^{225}}{1-(-3)})=4(\frac{1-(r)^{m+1}}{1-r})

Simplify.

-1(1-3^{225})+\frac{1+3^{225}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\-1+3^{225}+\frac{1}{2}+\frac{3^{225}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\-\frac{1}{2}+\frac{3(3^{225})}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\\frac{-1+3(3^{225})}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\\frac{-1+3^{226}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{-1+3^{226}}{8}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{1-3^{226}}{-8}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{1-9^{113}}{1-9}=4(\frac{1-(r)^{m+1}}{1-r})

Therefore, r = 9 and m = 112.

8 0
1 year ago
Albert thought of a number, added 5, multiplied the result by 2, took away 6 and then divided by 2 to give an answer of 8
ValentinkaMS [17]
The answer is 6 ....you just have to do the steps backwards and it will result faster.
6+5=11
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4 0
2 years ago
Read 2 more answers
Look at the middle 7 in the decimal 0.777. How is its value related to the value of the 7 to its left? To the value of the 7 to
masya89 [10]

Answer:

Value of middle 7 = \frac{1}{10}(value of 7 to the left of the middle 7)

Value of middle 7 = 10(value of 7 to the right of the middle 7)

Step-by-step explanation:

7 in the middle of number 0.777 is equal to \frac{7}{100}

7 to the left of the middle 7 is equal to \frac{7}{10}

So,

\frac{7}{100}=(\frac{1}{10}) \frac{7}{10}

That is value of middle 7 = \frac{1}{10}(value of 7 to the left of the middle 7)

Also,

7 to the right of the middle 7 is equal to \frac{7}{1000}

So,

\frac{7}{100}=(10) \frac{7}{1000}

That is value of middle 7 = 10(value of 7 to the right of the middle 7)

5 0
1 year ago
The value of x must be greater than _____
bezimeni [28]
Answer: The value of x must be greater than 3

------------------------------------------------------------------------
------------------------------------------------------------------------

Explanation:

The triangle inequality theorem will come into play here. The basic idea of this theorem is "take any two sides of a triangle and add them up. That sum must be larger than the third side". 

So if we have 
a = 12
b = x
c = 15

then the following must hold true (all three inequalities must be true)
a+b > c
a+c > b
b+c > a

Focus on the first inequality and plug in the given values. Then solve for x
a+b > c
12+x > 15
12+x-12 > 15-12
x > 3

So we see that x > 3. Repeat the same for the second inequality
a+c > b
12+15 > x
27 > x
x < 27

Repeat again for the third inequality
b+c > a
x+15 > 12
x+15-15 > 12-15
x > -3

-------------------------

In summary so far, we have
x > 3
x < 27
x > -3

Combine all of those inequalities to form one single compound inequality which is 3 < x < 27

For some reason your teacher doesn't want you to focus on the "less than 27" part, so it seems like s/he only wants the "x > 3" portion.

So this is why x must be larger than 3 (up only til you get to 27 though).

8 0
1 year ago
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