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____ [38]
4 years ago
7

What point lies on the line of the equation y+1=-0.6(x+4)

Mathematics
1 answer:
noname [10]4 years ago
4 0
The slope is: -0.6
The Y-Intercept is: -3.4
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Answer:

4b + 4 = 58

Step-by-step explanation:

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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

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Answer:

-72rs

Step-by-step explanation:

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