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zaharov [31]
2 years ago
13

2 factors that add to 6 but multiply to 60

Mathematics
1 answer:
hodyreva [135]2 years ago
8 0

Answer:

The short answer is there isn’t.

Start by writing each of these as an expression:

x * y = 60

x + y = 7

Next, solve each for the same variable; in this case, y:

(x * y) / x = 60 / x

.: y = 60 / x

(x + y) - x = 7 - x

.: y = 7 - x

Next, replace y of the second expression to the first

y = 60 / x & y = 7 - x

.: 7 - x = 60 / x

Now, solve for x:

(7 - x) * x = (60 / x) * x

.: x * 7 - x^2 = 60

This is quadratic, so write it in the form of ax2 + bx + x = 0

(-1)x^2 + (7)x + (-60) = 0

.: a = -1, b = 7, c = -60

Finally solve for b:

x = (-b +- sqrt(b^2 - 4*a*c)) / 2a

.: x = (-7 +- sqrt(7^2 - 4*-1*-60)) / (2 * -1)

.: x = (-7 +- sqrt(49 - 240)) / -2

.: x = (-7 +- sqrt(-191)) / -2

The square root of a negative value is imaginary and thus there’s no real answer to this problem.

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Rufina [12.5K]

Answer:

X = 4

Step-by-step explanation:

In the equation add 3 to both sides of the equation

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1/2x + 3/4x = -3/4x + 5 + 3/4x

Simplify

5/4x=5 Multiply both sides by 5

5x=20 divide both sides by 5

5x divided by 5 and 20 divided by 5

=4

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3 years ago
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Answer:

b

Step-by-step explanation:

4 0
2 years ago
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

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5 0
1 year ago
Find the equation of the tangent line at the point (1, 6). for y = 4 + 4x2 - 2x3.
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y = 4 + 4 x^{2}-2x^{3}  Lets \ write \ it \ y = -2 x^{3} +4 x^{2} +4&#10;
The tangential line at a certain point is just the derivative so.
y ' = -6 x^{2} +8x. At the point (1,6) we plug the x value in and get the slope at the point (y ' = 2)

The tangential line at that point is
y - 6 = 2(x - 1)     (this is the answer)

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