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puteri [66]
3 years ago
8

Please solve the equation with a rational coefficients,

Mathematics
1 answer:
Elodia [21]3 years ago
5 0
  • <em>Answer:</em>

<em>m = - 1.5</em>

  • <em>Step-by-step explanation:</em>

<em>Hi there !</em>

<em>- 1.4m = 2.1</em>

<em>m = 2.1 : - 1/4</em>

<em>m = - 1.5</em>

<em>Good luck !</em>

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Write an equation of the line in slope-intercept form given the slope is 2 and the y-intercept is -5.
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Answer:

y=2x-5

Step-by-step explanation:

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3 years ago
Given f(x)=9x-7, g(x)= 7x-6 find (f-g)(x)
lina2011 [118]
Answer: 2x - 1
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9x - 7 - 7x + 6
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5 0
2 years ago
The t value for a 95% confidence interval estimation with 24 degrees of freedom is
Alborosie

Answer: the value us 2.064

Step-by-step explanation:

To determine the t value, we would need the t distribution table. Since degree of freedom is known as 24, we would determine alpha/2

A 95% confidence level is 95/100 = 0.95

1 - alpha = 0.95

alpha = 1 - 0.95

alpha = 0.05

alpha/2 = 0.05/2 = 0.025

this is the area to the left. The area to the right 1 - 0.025 = 0.975

alpha/2 = 0.975

Looking at the t distribution table, the t value is

2.064

4 0
3 years ago
Write down the first term in the sequence given by: T(n)=n^2+3
Anton [14]

Some people call 'n' "zero" for the first term, others call 'n' "1".

-- If n=0 for the first term, then T(0) =  0² + 3  =  3 .

-- If n=1 for the first term, then  T(1) = 1² + 3  =  4 .
8 0
3 years ago
Read 2 more answers
4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia
Ad libitum [116K]

The <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its <em>standard</em> form from the coordinates of the foci and a <em>general</em> equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the <em>general</em> equation of the circle into its <em>standard</em> form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0​

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

#SPJ1

5 0
2 years ago
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