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Ghella [55]
3 years ago
8

PLEASE HELP ME I DONT UNDERSTAND

Mathematics
1 answer:
max2010maxim [7]3 years ago
5 0

Answer:

no

Step-by-step explanation:

the square root of 2 is not rational

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The model represents a polynomial of the form ax2 + bx + c. Which equation is represented by the model?
Maru [420]

Consider the given table.

  • The first row represents first term (first brackets):

                                +x+x+x-1=3x-1

  • The first column represents second term (second brackets):

                                +x-1=x-1

  • The product of (3x-1) and (x-1) is equal to the sum of all remaining terms in this table:

                               +x²+x²+x²-x-x-x-x+1=3x²-4x+1.

Therefore, 3x²-4x+1=(3x-1)(x-1).

Answer: correct choice is C.

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A stock market analyst examined the prospects of the shares of a large number of corporations. When the performance of these sto
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Answer:

Solution:Bayes:E1: Stock performs much better than the market averageE2: Stock performs same as the market averageE3: Stock performs worse than the market averageA: Stock is rated a ‘Good Buy’Given thatP(E1) = .25,P(E2) = .5,P(E3) = .25,P(A| E1) = .4,P(A| E2) = .2,P(A| E3) = .1Then,

P A EP EP EAP A EP EP A EP EP A EP E=++=(.40)(.25).444(.4)(.25)(.2)(.5)(.1)(.25)

8 0
3 years ago
Which is the slope of a line perpendicular to the line 7x + y equals 9​
Neporo4naja [7]

Answer:

slope = 1/7

Step-by-step explanation:

7x+y=9 you should be in slope int form--> y=-7x+9

then the line is perpendicular so

the slope is 1/7

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3 years ago
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Given f(x)=2x^2+3x-5 for what values of x is f(x) positive
kow [346]

Answer:

The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)

Step-by-step explanation:

we have

f(x)=2x^{2}+3x-5

This is a vertical parabola open upward (the leading coefficient is positive)

The vertex is a minimum

The coordinates of the vertex is the point (h,k)

step 1

Find the vertex of the quadratic function

Factor the leading coefficient 2

f(x)=2(x^{2}+\frac{3}{2}x)-5

Complete the square

f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-5-\frac{9}{8}

f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-\frac{49}{8}

Rewrite as perfect squares

f(x)=2(x+\frac{3}{4})^{2}-\frac{49}{8}

The vertex is the point (-\frac{3}{4},-\frac{49}{8})

step 2

Find the x-intercepts (values of x when the value of f(x) is equal to zero)

For f(x)=0

2(x+\frac{3}{4})^{2}-\frac{49}{8}=0

2(x+\frac{3}{4})^{2}=\frac{49}{8}

(x+\frac{3}{4})^{2}=\frac{49}{16}

take the square root both sides

x+\frac{3}{4}=\pm\frac{7}{4}

x=-\frac{3}{4}\pm\frac{7}{4}

x_1=-\frac{3}{4}+\frac{7}{4}=1

x_2=-\frac{3}{4}-\frac{7}{4}=-2.5

therefore

The function f(x) is negative in the interval (-2.5,1)

The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)

see the attached figure to better understand the problem

6 0
4 years ago
Please help me with these ​
kumpel [21]

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