According to the number line, which statement MUST be true?

The sum of three consecutive even integers is 258. write an equation ad solve to find the integers

slavikrds [6]

The consecutive integers are 85,86,87

Explanation:

n

:

the first number

n

+

1

:

the second number

n

+

2

:

the third number

n

+

(

n

+

1

)

+

(

n

+

2

)

=

258

3

n

+

3

=

258

3

n

=

258

−

3

n

=

255

n

=

255

3

n

=

85

n

+

1

=

85

+

1

=

86

n

+

2

=

85

+

2

=

87

8 0

3 years ago

Read 2 more answers

Deena has 4 pairs of of white socks , 3 pairs of black sock, 1 pair of red socks, and 2 pairs of Navy socks in her drawer each p

VikaD [51]

Answer:

2/5

Step-by-step explanation:

Ok there are 4+3+1+2 pairs of socks in all... Aka 10 sock pairs in all.

There are 4 white pairs. So the probability she selects a white pair is 4/10.

4/10 can be reduced to 2/5.

7 0

2 years ago

Read 2 more answers

The fish population in a pond evolves according to the logistic equation and a percentage of fish are removed for each unitof ti

musickatia [10]

Step-by-step explanation:

Suppose that the carrying capacity is 900 fish, the intrinsic growth rate is equal to 2, and the harvest rate is 20%.

a. Use the graphical or eigenvalue approach

8 0

3 years ago

A football is selling for $35 off the original price. The original price was $60. What is the sale price of the football?

Strike441 [17]

All you have to do it do 60 - 35 = 25 dollars because they want to know if the original price would be 60 dollars and they wanna know the anwser if you took away 35 dollars. so then you just subtract

5 0

3 years ago

John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m

Dmitry [639]

Answer:

The minimum possible initial amount of fish: $52$

Step-by-step explanation:

Let's start by saying that

$x$ = is the initial number of fishes

John:

When John arrives:

he throws away one fish from the bunch

$x-1$

divides the remaining fish into three.

$\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}$

takes a third for himself.

$\dfrac{x-1}{3} + \dfrac{x-1}{3}$

the remaining fish are expressed by the above expression. Let's call it John

$\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}$

and simplify it!

$\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}$

When Joe arrives:

he throws away one fish from the remaining bunch

$\text{John} -1$

divides the remaining fish into three

$\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}$

takes a third for himself.

$\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

the remaining fish are expressed by the above expression. Let's call it Joe

$\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

and simiplify it

$\text{Joe}=\dfrac{2}{3}(\text{John}-1)$

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

$\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)$

$\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}$

When James arrives:

We're gonna do this one quickly, since its the same process all over again

$\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}$

$\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)$

$\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}$

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say $n$ ), a third is taken away by James. the remaining bunch would be $\frac{n}{3}+\frac{n}{3}$

hence we've expressed the last pile in terms of n as well. Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

$\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

$\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

$x=52$

This is minimum possible amount of fish before John threw out the first fish

8 0

2 years ago