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igor_vitrenko [27]
2 years ago
7

Find the equation of the line with slope -2 which goes through the point (9,-7).

Mathematics
1 answer:
ryzh [129]2 years ago
4 0
The answer is y= -2x-7
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Label x-intercepts, vertex and axis of symmetry for f(x) = (x+3)(x-5)
Lera25 [3.4K]
X intercepts are -3 and 5
7 0
3 years ago
How do I solve number one so I can do the rest?
Ipatiy [6.2K]

Answer:

multiply wtl

Step-by-step explanation:

4 0
2 years ago
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El Club de Matemáticas está horneando pasteles para una venta de
SVETLANKA909090 [29]

Answer:

56

Step-by-step explanation:

El problema se puede transcribir en esta ecuación:

2x + x = 168

siendo x las nectarinas

sumas los términos de x:

3x=168

despejas x:

x = 168 ÷ 3

x = 56

5 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
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
1 year ago
The range of the function f(k) = k2 + 2k + 1 is {25, 64}. What is the function’s domain? {5, 8} {-5, -8} {3, 8} {4, 7} {4, 8}
Savatey [412]
The range is the ouutput from inputing the input
basically
25=k²+2k+1 and 64=k²+2k+1
the values that satisfy both equations (not at the same tim) are the valuess that are the domain
solve each
25=k²+2k+1
minus 25 both sides (or recognize the perfect square trinomial, but anyway)
0=k²+2k-24
factor
0=(k+6)(k-4)
set to zero
k+6=0
k=-6
k-4=0
k=4
k=-6 or 4

64=k²+2k+1
minus 64 both sides
0=k²+2k-63
facor
0=(k-7)(k+9)
set to zer
k-7=0
k=7
k+9=0
k=-9
k=-9 or 7



so the domain has the numbers
-9,-6,4,7
it seems we only want the positive square roots so
answer is {4,7} is the domain
3 0
3 years ago
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