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4 years ago

The left and right page numbers of an open book are two consecutive integers whose sum is 385. Find these page numbers.

1 answer:

6 0

Consecutive integers are integers that follow one another. For example, 2,3,4,5,etc.... are consecutive integers

So algebraically, consecutive integers follow the form x, x+1, x+2, etc...

Since the sum of two consecutive integers is 239, this means:

x%2Bx%2B1=239

2x%2B1=239 Combine like terms on the left side

2x=239-1Subtract 1 from both sides

2x=238 Combine like terms on the right side

x=%28238%29%2F%282%29 Divide both sides by 2 to isolate x

x=119 Divide

So our first number is x=119

So to find the next number, simply add 1 to it to get 119%2B1=120

Answer: So our two page numbers are 119 and 120

You might be interested in

Suppose we have 14 red balls and 14 green balls as in the previous exercise. Show that at least two pairs, consisting of one red

Nuetrik [128]

Answer:

since each ball has a different number and if no two pairs have the same value there is going to be 14∗14 different sums. Looking at the numbers 1 through 100 the highest sum is 199 and lowest is 3, giving 197 possible sums

For the 14 case, we show that there exist at least one number from set {3,4,5,...,17} is not obtainable and at least one number from set {199,198,...,185} is not obtainable.

So we are left with 197 - 195 options

14 x 14 = 196

196 > 195

so there are two pairs consisting of one red and one green ball that have the same value

As to the comment, I constructed a counter-example list for the 13 case as follows. The idea of constructing this list is similar to the proof for the 14 case.

Red: (1,9,16,23,30,37,44,51,58,65,72,79,86)

Green: (2,3,4,5,6,7,8;94,95,96,97,98,99,100)

Note that 86+8=94 and 1+94=95 so there are no duplicated sum

Step-by-step explanation:

For the 14 case, we show that there exist at least one number from set {3,4,5,...,17} is not obtainable and at least one number from set {199,198,...,185} is not obtainable.

First consider the set {3,4,5,...,17}.

Suppose all numbers in this set are obtainable.

Then since 3 is obtainable, 1 and 2 are of different color. Then since 4 is obtainable, 1 and 3 are of different color. Now suppose 1 is of one color and 2,3,...,n−1 where n−1<17 are of the same color that is different from 1's color, then if n<17 in order for n+1 to be obtainable n and 1 must be of different color so 2,3,...,n are of same color. Hence by induction for all n<17, 2,3,...,n must be of same color. However this means there are 16−2+1=15 balls of the color contradiction.

Hence there exist at least one number in the set not obtainable.

We can use a similar argument to show if all elements in {199,198,...,185} are obtainable then 99,98,...,85 must all be of the same color which means there are 15 balls of the color contradiction so there are at least one number not obtainable as well.

Now we have only 195 choices left and 196>195 so identical sum must appear

A similar argument can be held for the case of 13 red balls and 14 green balls

6 0

4 years ago

A homing pigeon flew a distance of 120 miles in 4 days. If ti flew x miles in the first 3 days, write an expression to tell how

Diano4ka-milaya [45]

×120 and 3 then what ever you got for 120×3 ×that number with 4

3 0

4 years ago

How to write 8 17/20 as a decimal number

worty [1.4K]

Hello!

Answer:

In order to convert mixed number 8 17/20 to decimal form, you follow these steps:

Convert the proper fraction to a decimal: Divide the numerator by the denominator

17 ÷ 20 = 0.85

Add this decimal number to the whole number of the mixed fraction

8 + 0.85 = 8.85

The mixed number 8 17/20 converted to decimal number is, therefore:

8.85

Hope this helps <3

4 0

3 years ago

Read 2 more answers

I need to solve these problems. Plz solve theses problems ASAP.

podryga [215]

Answer:

Step-by-step explanation:

Let's start with the first problem :

Dora spent 1/3 of her money and it equals 72
Jack spen 1/4 and had the amount left

To get the initial amount that Dora had we must multliply 72 by 3 since she spent 1/3 and 3 is the total number of parts

72*3=216 ⇒ Dora had 216
she spent 72 so : 216-72=144 ⇒144 is the amount left for her and Jack

Jack spent 1/4 of his total amount and had 144 left

144 is 3/4
to get 1/4 we must divide 144 by 3 ⇒ 144/3=48
so the total amount is 48*4=192

Let's check : 192-48= 144 so it's right

Jack spent 48

The second question :

We notice that the total area A is the sum of the white square's area and the blue region's area

A= 65+ x x is the area of the white square so x = 4*4 since it is 4cm*4cm
so A= 65+16= 81
we khow that the area of a square is the lenght times itself
so 81 = y² with y the lenght of the large blue square and y≥0 since it is a distance
y= $\sqrt{81}$ = 9
so the lenght is 9 cm