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viktelen [127]
3 years ago
10

Determine the intervals on which the polynomial is entirely negative and those on which it is entirely positive. (Enter your ans

wers using interval notation. Enter EMPTY or ∅ for the empty set.)
−x^2 + 6x − 10
I don't understand how to factor this, as as such, I can't figure out how to answer the question. Please help!
Mathematics
2 answers:
Stells [14]3 years ago
5 0
This is a parabola that opens downward.
Remember that the axis of symmetry is located at x=-b/(2a).
So the vertex is at (3,-1)
So functions slope is positive or increasing on (-co, 3] and decreasing on [3,co).

Law Incorporation [45]3 years ago
3 0

Answer with explanation:

The function for which we have to find , the intervals on which the polynomial is entirely negative and those on which it is entirely positive.

 f(x)= -x²+6 x -10

 If you will find the root of the function, there is no real root.

To find the root we will use Discriminant formula

For a Quadratic function, ax²+b x+c=0,

 x=\frac{-b+\sqrt{D}}{2a}\\\\D=b^2-4ac\\\\x=\frac{-6\pm \sqrt{6^2-4 *(-1)*(-10)}}{2*(-1)}\\\\x=\frac{-6\pm\sqrt{-4}}{-2}

→So,there is no interval in which  the polynomial is entirely negative and those on which it is entirely positive, which can be represented in interval notation using Ф.

   f(x)= -(x^2-6x+10)\\\\= - [(x-3)^2-9+10]\\\\=-(x-3)^2-1

For, any value of x,the value of f(x) will be always Negative.

To find the vertex, put ,x-3=0

 x=3

And, by putting , x= 3 ,in the equation we get

y = -1

So,Vertex = (3, -1)

⇒If you will try to find the intervals in which the function is increasing , means the curve is moving up is from (-∞, 3) and the intervals in which the function is decreasing , means the curve is moving downward is from , (3, ∞).

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Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

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move 3x to the other side

5y=-3x+15

divide by 5

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now, perendicular lines have slopes that are negative and reciprocal.

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2=5/3(3)+b

2=5+b

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so that means that our equation will now be

y=5/3x-3.

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