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ValentinkaMS [17]
3 years ago
15

Enter the output values (Y) with the given function and input values (X). y = 3x - 2

Mathematics
2 answers:
svlad2 [7]3 years ago
6 0
6=16
5=13
4=10
7=19
i might be wrong not sure tho
marishachu [46]3 years ago
5 0

Answer:

hilghkiii

Step-by-step explanation:

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Need Help Finding average rate of change on a graph? I have a screenshot of the graph and the question
Katen [24]

Answer:

-3

Step-by-step explanation:

The average rate of change is the slope of the line that connects the points of interest. At x=-6, the point is (-6, 8). At x=-5, the point is (-5, 5).

I find it easy to refer to the graph to see that the point at x=-5 is 3 units down and 1 unit to the right of the point at x=-6. So the slope of the connecting line is -3/1 = -3.

If you want to find the same result analytically, you can compute ...

... ∆y/∆x = (y2 -y1)/(x2 -x1)

... = (5 -8)/(-5 -(-6))

... = -3/1 = -3

8 0
3 years ago
The University of South Texas is interested in determining if there is a difference in the average weight gain between male stud
Blizzard [7]

Answer: A. We are 95% confident that the population mean weight gain for male University of South Texas students is 2.4 more to 20,4 more than female University of South Texas students.

Step-by-step explanation:

Given a 95% confidence interval of : (2.4, 20.4)

The lower boundary of the interval = 2.4

The upper boundary of the interval = 20.4

Male = group 2, Female = group 1

P2 - P1 is between lower boundary more and upper boundary more ;

Hence

Population mean weight for male is 2.4 more to 20.4 more than female

6 0
3 years ago
(cos x - (sqrt 2)/2)(sec x -1)=0
Darya [45]

(\cos x-\frac{\sqrt{2}}{2})(\sec x-1)=0 [/tex]

=(\cos x-\frac{1}{\sqrt{2}})(\sec x-1)=0 [/tex]

\frac{(\sqrt{2}\cos x-1)}{\sqrt{2}}(\frac{1}{\cos x\ }-1)=0

(Reciprocal Identity)

(\frac{^{\sqrt{2}\\cos  x-1}}{^{\sqrt{2}}})(\frac{1-\cos x}{\cos x})=0

\frac{^{(\sqrt{2}\cos x-1})}{\sqrt{2}}\frac{(1-\cos x)}{\cos x}=0

(\sqrt{2}\cos x-1}){(1-\cos x)}=0 (ZeroProduct Property)

\sqrt{2}\cos x-1=0

\sqrt{2}\cos x=1

\cos x=\frac{1}{\sqrt{2}}

x=\frac{\Pi }{4}

and

1-\cos x=0

\cos x=1

x=0

x=0 and x=\frac{\Pi }{4} are the solutions.

6 0
4 years ago
Read 2 more answers
In the equation x^2+kx+54=0 one root is twice the other root find k
Maurinko [17]
Let r be the lesser root and r^2 be the greater.......the sum of the roots   = -b/a  = -[-6] / 1  = 6

 

So we have that

 

r^2 + r  = 6     →    r^2 + r - 6 = 0      

 

Factor

 

(r + 3) (r - 2)   = 0  

 

So r = -3    or   r = 2

 

Then r^2  = 9     or r^2  =4

 

And the product of the roots =  c/a   =  k/a  = k

 

So....k =   (-3)(9)   = -27     or k = (2)(4)  = 8

 

Check

 

x^2 - 6x - 27  = 0      factors as  (x + 3)(x - 9) = 0     and  the roots are -3 and 9

 

x^2 - 6x + 8  = 0      factors as (x -2) (x - 4) =0      and the roots are 2 and 4

 

6 0
3 years ago
A car rental agency charges $10 to reserve a car and $25 per day. You have $160 to spend on the rental car.
ira [324]
You would have a maximum of 6 days
$160-$10 = $150
$150/$25=$6
6 0
3 years ago
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