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maks197457 [2]
3 years ago
13

Solve the equation -7 (8a - 9 = - 23 + 4a

Mathematics
1 answer:
lapo4ka [179]3 years ago
8 0

Answer:

a= 1.5

Step-by-step explanation:

distrubutive property:

-7(8a) - 7(9) = -23 +4a

-56a - 63 = -23 + 4a

+63 +63

-56a = 40 + 4a

-4a -4a

-60a=40

a= -1.5

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Find the quotient 12÷3/8
bearhunter [10]

Answer:

32

Step-by-step explanation:

To make it a fraction form answer, you multiply the whole number by the denominator and make the result the new numerator. The old numerator becomes the new denominator

 

Thus, the answer to 12 divided by 3/8 in fraction form is:

96 /3

To make the answer to 12 divided by 3/8 in decimal form, you simply divide the numerator by the denominator from the fraction answer above:

96 / 3 = 32

The answer is rounded to the nearest two decimal points if necessary.

96/3 can be simplified to 32/1.

32/1 is an improper fraction and should be written as 32.

Hope this helps <3

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They are all larger than 90 degrees, and in a triangle they would all be obtuse angles.
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In a carnival game, you roll 2 dice. if the sum is 6, you receive a $5 payoff. if the sum is 12, you receive a $9 payoff. otherw
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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
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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
1 year ago
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