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soldi70 [24.7K]
3 years ago
5

For which value of m does the graph of y = 18x2 + mx + 2 have exactly one x-intercept?

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
4 0

Answer:

m = +/- 12

Step-by-step explanation:

A quadratic function has only one x-intercept when the discriminant is equal to 0. The discriminant is b^2 - 4(a)(c).

When we plug in what we know, we have:

m^2 - 4(18)(2) = 0.

Then using algebraic properties, solve for m.

m^2 - 144 = 0

m^2 = 144

m = +/- 12

When you plug in positive or negative 12, and then factor you will see that it comes out to a difference of squares, proving that there is only one x-intercept.

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I don’t understand please help!
nikklg [1K]

Answer:

x > 10/3

x will always be greater that 10/3 never the same nor less.

Step-by-step explanation:

3/5x + 1/3 < 4/5x - 1/3

3/5x - 4/5x + 1/3 < -1/3

-1/5x + 1/3 < -1/3

-1/5x < -1/3 - 1/3

-1/5x < -2/3

-1/5x/-1 < -2/3/-1   (Since it is both divided by a negative the sign switches, the sign also switches with multiplication. *Note: it only switches if it is negative!*)

1/5x > 2/3

5 • 1/5x > 2/3 • 5

x > 2/3 • 5/1

x > 10/3

6 0
3 years ago
The money remaining from the $150 is 37.5% of the cost of the day trip to cairo
weeeeeb [17]
The cost of the trip was $56.25
5 0
3 years ago
Write 4/28 in simplest form
77julia77 [94]
Simplists form is 1/7


7 0
3 years ago
Read 2 more answers
2 3/6 boxes and 1 2/3 boxes were recycled what is the total
Liono4ka [1.6K]
Since you are recycling the boxes you will be subtracting 2 3/6 and 1 2/3 since you’re a denominators you’re not mad she have to find a common denominator which would be six. So you have to do 2/2 x 2/3 = 4/6. So then you would do 3/6 - 4/6 = 1/6. Then you would subtract 1-2=1. Then you would do 1 + 1/6 = 1 1/6

So the answer is 1 1/6.
I hope this helped
5 0
3 years ago
What are the coordinates of the inflection point on the graph of y=(x+1)arctanx
Dennis_Churaev [7]

Inflection point is the point where the second derivative of a graph is zero.

y = (x+1)arctan xy' = (x+1)(arctan x)' + (1)arctan xy' = (x+1)/(x^2+1) + arctan xy'' = (x+1)(1/(1+x^2))' + 1/(1+x^2) + 1/(1+x^2)y'' = (x+1)(-1/(1+x^2)^2)(2x)+2/(1+x^2)y'' = ((x+1)(-2x)+1+x^2)/(1+x^2)^2y'' = (-2x^2-2x+2+2x^2)/(1+x^2)^2y'' = (-2x+2)/(1+x^2)^2

Solving for point of inflection: y'' = 00 = (-2x+2)/(1+x^2)^20 = -2x+2x = 1y(1) = (1+1)arctan(1) = 2 * pi/4 = pi/2

Therefore, E(1, pi/2).


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0
3 years ago
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