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lesya692 [45]
3 years ago
13

The number 2017 can be expressed as the difference of the squares of two consecutive whole numbers. What is the sum of these two

numbers?
Mathematics
1 answer:
SSSSS [86.1K]3 years ago
8 0

Let x be the first number

X + 1 be the second number

So

(x+1)^2 – x^2 = 2017

X^2 + 2x +1 – x^2 = 2017

2x + 1 = 2017

2x = 2017 – 1

2x = 2016

X = 1008

X + 1 = 1009

Sum = 1008 + 1009 = 2017

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The Boy Scouts self popcorn Ethan sold 9 small
NARA [144]

Answer:

The cost of 1 small bag of popcorn = $2.95

The cost of 1 large bag of popcorn   = $4.75

Step-by-step explanation:

Here, let us assume the cost of 1 small bag of popcorn = $ m

And,  the cost of 1 large bag of popcorn = $ n

So, the cost of 9 small bag = 9 x ( cost of 1 small bag)  = 9 x (m) = 9 m

and  the cost of 2 large bag = 2 x ( cost of 1 large bag)  = 2 x (n) = 2 n

Also, cost of 9 small + 2 large bag of popcorn sold by  Ethan for   $ 36.05 .

⇒ Total cost of 9 small + 2 large bag of popcorn =  9 m + 2 n

⇒ 9  m +  2 n =  $ 36.05  ..........  (1)

Also, the cost of 5 small bag = 5 x ( cost of 1 small bag)  = 5 m

and  the cost of 7 large bag = 7 x ( cost of 1 large bag)  = 7 n

Also, cost of 5 small + 7 large bag of popcorn sold by  Jack for   $ 48 .

⇒ Total cost of 5 small + 7 large bag of popcorn =  5 m + 7 n

⇒ 5 m +  7 n =  $ 48 ..........  (2)

Now, solving (1) and (2) for the values of m and n , we get

9  m +  2 n =   36.05

5 m +  7 n =  48

Multiply (1) with (-7) and (2) by 2 , add both equation we get:

-63 m - 14 n  + 10 m + 14 n =  -252.35 + 96

or, - 53 m = -156.35

or, m =   2.95

Now, put the value of m in (1), we get:  5 ( 2.95) + 7 n =  48

or, n =  33.25 /7 = 4.75

Hence the cost of 1 small bag of popcorn = $2.95

The cost of 1 large bag  = $4.75

6 0
3 years ago
Read 2 more answers
A child has a box of candies which might have a toy inside . The odds against the box having toys are 17/2 . What is the probabi
Elden [556K]
17/2 = probability is low
4 0
3 years ago
In mom's car, the number of miles driven is proportinal to the number of gallons of gas used.
DerKrebs [107]
For number 1 the answer is 112

7 0
3 years ago
Solve the quadratic equation numerically (using tables of x- and y- values). -4x=-x2+12
Levart [38]
-4x=-x2+12 should be written  <span>-4x=-x^2+12, with " ^ " representing exponentiation.

You are to invent a table of x values and find the difference between -4x and -x^ 2+12.  The smaller the difference, the closer you are to finding the correct root, x.

To get you started:

 x     -4x              x^2+12      difference
--- ------------   ----------------   --------------
 2       -8                 16               24 (pretty bad!)
-4       16                28              12 (better)
 -10   40              112                72 (much worse)

To answer this problem correctly, you must present such a table, even tho' it'd be faster to use the quadratic formula (or some other method) to find x.


</span>
5 0
3 years ago
Please help me! I’m failing!!!
choli [55]

a. If G(t) = a (1 + r)ᵗ, then G(0) = a corresponds to the starting area of the glacier, which is given to be 142 acres. So a = 142.

The area of the glacier shrinks by 4.4% each year. This means that after 1 year, the area is reduced to

142 - (4.4% of 142) = 142 (1 - 0.044) ≈ 135.75 acres

After another year,

135.75 - (4.4% of 135.75) = 135.75 (1 - 0.044) ≈ 129.78 acres

or equivalently, 142 (1 - 0.044)² acres.

And so on. After t years, the glacier would have an area of

G(t) = 142 (1 - 0.044)ᵗ   ⇒   G(t) = 142 × 0.956ᵗ

b. If the year 2007 corresponds to t = 0, then 2012 refers to t = 5. Then the area of the glacier is

G(5) = 142 × 0.956⁵   ⇒   G(5) ≈ 113.39 acres

c. Simply take the difference between G(5) and G(0) :

G(5) - G(0) ≈ 142 - 113.39 ≈ 28.61 acres

d. The average rate of change of G(t) from 2007 to 2012 is given by the difference quotient of G(t) over the interval 0 ≤ t ≤ 5 :

ARC_{[2007,2012]} = \dfrac{G(5) - G(0)}{5 - 0} \approx \dfrac{113.39 - 142}5

so that ARC ≈ -5.72 acres/year. This rate tells us that a little less than 6 acres of glacial ice is lost each year, based on the reduction over the first 5 years.

e. The year 2017 refers to t = 10, so now we compute

ARC_{[2012,2017]} = \dfrac{G(10) - G(5)}{10 - 5} \approx \dfrac{90.55-113.39}5

which comes out to about ARC ≈ -4.57 acres/year. Since this rate is smaller than the ARC between 2007 and 2012, this means that the glacial ice is disappearing at a slower rate in later years.

8 0
2 years ago
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