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Novosadov [1.4K]
4 years ago
13

How to solve

{y}(5y) = 2" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Natalija [7]4 years ago
5 0

Answer:

  y = 5

Step-by-step explanation:

Expand the logarithm:

\log_y{(5)}+\log_y{(y)}=2\\\\\dfrac{\log{(5)}}{\log{(y)}}+1=2 \quad\text{change of base formula}\\\\\dfrac{\log{(5)}}{\log{(y)}}=1 \quad\text{subtract 1}\\\\\log{(5)}=\log{(y)} \quad\text{multiply by log(y)}\\\\5=y \quad\text{take the anti-log}

_____

You can also take the antilog first:

  5y = y²

  y(y -5) = 0 . . . . . subtract 5y, factor

  y = 0 or 5 . . . . . y=0 is not a viable solution, so y=5.

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What is the slope of the linear function 5x + y = 9?<br> a. 1<br> b. -5<br> c. 5<br> d. 9
Novay_Z [31]

Answer:

b. because m = -5

Step-by-step explanation:

y = -5x +9

m (gradient) is the coefficient before the 'x' term

7 0
2 years ago
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Nick can read 3 pages in 1 minute. write the ordered pairs for nick reading 0,1,2 and 3 min
Elis [28]
For 0 minutes, it would be 0 minutes (obviously)
for 1 minute, it would be 3
for 2 minutes, it would be 6 pages
and finally, 3 minutes, will be 9.
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4 0
4 years ago
4 students measure their height. Their heights measure 184 cm, 192 cm, 164 cm, and 200 cm. What is the mean height of the studen
melomori [17]
The mean is the average, which is the sum of all the numbers divided by the total numbers there are. I will add them up for you, and then work from there.
184 + 192 + 164 + 200 = 740.
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740/4 = 185.
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I hope this helps!
8 0
3 years ago
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Which point is on the graph of f(x) = 2 • 5x?
Inessa [10]

Answer:

1, 10

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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
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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