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To find the portion of the class that did not study, we can set up an equation using the given information: Let x be the proportion of students who studied and (1 - x) be the proportion of students who did not study. The equation can be set up as follows: \(78x + (1 - x) 54 = 70\) . Solving this equation, we get:

1. \(78x + 54 - 54x = 70\)
2. \(24x + 54 = 70\)
3. \(24x = 16\)
4. \(x = 16 / 24\)
5. \(x = 2 / 3\)

Therefore, the portion of the class that did not study is \(1 - 2/3 = 1/3\).

Answer:

n = 1

Explanation:

    You want the largest positive integer n such that n³ divides 15.

Divisors

The positive integer divisors of 15 are ...

  1, 3, 5, 15

The only one of these that is a cube is 1 = 1³.

  n = 1

 Hope u found this helpful!

Brandon spends his afternoon picking apples from an orchard.  He notices that if he groups the apples he picked by fives, six, or sevens, then he ends up with $1$ left over.  If he groups by eights, he has $1$ left over.  If Brandon knows that he picked between $1000$ and $2000$ apples, how many did he pick?   

I multiplied 5 • 6 • 7 • 8 and got 1680.  Then I added 1 to it.   

                                              Brandon picked 1681 apples   

check answer   

                                              1681 / 5  =  335 R 1   

                                              1681 / 6  =  280 R 1   

                                              1681 / 7  =  240 R 1   

                                              1681 / 8  =  210 R 1   

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Find the sum of all values of $k$ for which $x^2 + kx - 9x + 25 - 9$ is the square of a binomial.   

                                                                       x² + (kx – 9x) + 16   

factor out x                                                      x² + (k–9)(x) + 16

When posed as ax² + bx + c, for there to be     

a double root, also called a repeated root,   

the term c must be a square and b = 2sqrt(c)  

Square root of 16 = +4 so b = (2)(+4) = +8                  

when +4 is positive                                         k – 9 = +8        

                                                                       k = +8 + 9   

                                                                       k = 17  

when +4 is negative                                        k – 9 = –8   

                                                                       k = –8 + 9   

                                                                       k = 1   

the sum of all values of k                                sum = 17 + 1   

                                                                       sum = 18   

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The minimum value is 2.

The answer is [0,infty).

The answer is 28.

There is no graph shown.  Relying only on the three points provided, one   

parabola that passes through the points (–2,8), (0,0), and (2,8) is y = 2x²    

If that is correct, then a + b + c  =  2 + 0 + 0  =  2.   

Those three points, and the vertex at (0,0), indicate that the parabola    

is symmetrical to the y-axis and that the equation has a double root.  

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1. first we simplfy 

y = -5x^2 -6x + 10 is the most simplfied version 

* We can now convert this to vertex form

y= -5 ( x^2 + 6/5 x - 2 ) 

==> y = -5 ( x + 3/5)^2 +59/5  is the vertex form

vertex ==> (h,k)  and the standard vertex form == >  y = a ( x − h ) 2 + k

thus vertex ==> (-3/5 , 59/5)

Yes, it is correct.  I was still typing when your answer was   

posted, but we both solved it the same way.  Good work.   

_,

Find the constant k such that the quadratic 2x^2 + 3x + 8x - x^2 + 4x + k has a double root.  

                                                                    2x² + 3x + 8x – x² + 4x + k   

combine like terms                                         x² + 15x + k   

for a quadratic to have a double root,   

which is also called a repeated root,   

its discriminant must equal zero

when a quadratic is in the form ax² + bx +c   

its discriminant is the quantity b² – 4ac            b² – 4ac   

                                                                       15² – (4)(1)(k)   

                                                                       225 – 4k    

solve for k                                                          k  =  (225 / 4)    

                                                                           k  =  56.25    

by the way, the quadratic in this problem factors to (x + 7.5)(x + 7.5)   

_.

for a quadratic to have a double root we want the variables of b^2 -4ac = 0

we can combine like terms in this quadratic via simplification and get 

x^2+11x+4x + k 

==> x^2 + 15x + k = 0

now we just plug this into the discriminant equation ( b^2 - 4ac = 0  ) 

== > 15^2 - 4 x 1 x c =0

simplfy == > 225 - 4c = 0 

thus c = 56.25

please check if this is right 

First, we expand the left-hand side of the equation: (2x + 7)(x - 4) = 2x^2 - 8x + 7x - 28 = 2x^2 - x - 28

Then, we rewrite the equation with all terms on one side: 2x^ 2- x - 28 = -30 + jx

Simplifying, we obtain: 2x^2 + (j - 1) x + 2 = 0

For the quadratic equation to have exactly one real solution, the discriminant must be zero. That is, b^2 - 4ac = 0

Substituting a = 2, b = j - 1, c = 2 into the discriminant formula, we get:

(j - 1)^2 - 4(2)(2) = 0

Solving for j, we find:

j^2 - 2j + 1 - 16 = 0

j^2 - 2j - 15 = 0

Factoring, we get:

(j - 5)(j + 3) = 0

Thus, the possible values of j are 5 and -3.

Let x be the number of tokens Eddie had at first.

Freddy received 3/10 * x tokens from Eddie.

Gary received 1/4 * x tokens from Eddie.

After giving away tokens, Eddie had x - 3/10x - 1/4x = 243 - x tokens left.

Multiplying both sides by 20 to get rid of the fractions, we get:

20x - 6x - 5x = 20 * 243 - 20x
9x = 4860 - 20x
29x = 4860
x = 4860 / 29
x = 168

So, Eddie had 168 tokens at first.

Freddy received 3/10 * 168 = 50 tokens from Eddie.

Gary received 1/4 * 168 = 42 tokens from Eddie.

Together, Freddy and Gary had 50 + 42 = 92 tokens at first.

Let's represent the initial amount of money each person had as follows:

Raju = R
Sam = S
Tristan = T

We know that:
R + S + T = 435

Raju spent 4/5 of his money, so he has 1/5 of his money left:
R = (1/5)R

Sam spent 2/3 of his money, so he has 1/3 of his money left:
S = (1/3)S

Tristan spent 3/4 of his money, so he has 1/4 of his money left:
T = (1/4)T

We also know that Raju has $55 more than Sam:
R = S + 55

And Tristan has $10 more than Sam:
T = S + 10

Now, we can substitute the expressions for R, S, and T into the total money equation:
(1/5)R + (1/3)S + (1/4)T = 435

Substitute R = S + 55 and T = S + 10 into the equation above:
(1/5)(S + 55) + (1/3)S + (1/4)(S + 10) = 435

Solving the equation above gives:
S = 180

Now that we have found the value of S, we can find the amount of money Tristan had at first:
T = S + 10
T = 180 + 10
T = 190

Therefore, Tristan originally had $190.

The number of pairs is 2.

The value of n is 11.

The solution is z = 2/5.

Let S be the number of students who studied and NS be the number of students who did not study

We can represent the total number of students (T) with the following equation

T = S + N

We are given the average scores for each group and the overall class average:

Average score for those who studied (AS): AS = 78

Average score for those who did not study (ANS): ANS = 54

Overall class average (Avg): Avg = 70

We can express the total score for all students using the average scores and the number of students in each group:

Total score for those who studied (TS): TS = AS * S

Total score for those who did not study (TNS): TNS = ANS * NS

Total score for the entire class (Total): Total = Avg * T (since Avg represents the average score per student)

Since the total score for the entire class is the sum of the scores from each group:

Total = TS + TNS

We can substitute the expressions for each score:

Avg * T = AS * S + ANS * NS

Now we can substitute the values we are given:

70 * (S + NS) = 78 * S + 54 * NS

We can rearrange the equation to isolate NS (the number of students who did not study):

20 * (S + NS) = 24 * S

20S + 20NS = 24S

-4S = 20NS

NS = (-4/20) * S (we can divide both sides by -4 as long as S is not 0, which we can safely assume since there must be at least one student who studied)

NS = -(1/5) * S

Since NS represents a number of students, it cannot be negative. Therefore, we can flip the signs to get a positive value for the fraction representing the portion of the class that did not study:

NS/T = (1/5) * S/T

We are interested in the portion of the class that did not study, so we can express this as:

Fraction of class who did not study = NS/T = 1/5.

If you let the first machine operate 2 times, it will have added 10 pieces,   

then you let the second machine operate 3 times, it will remove 9 pieces.   

This would leave 1 piece in the vat.   

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