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Furkat [3]
3 years ago
14

If y varies directly with x and y = 20 when x = 2,

Mathematics
1 answer:
Alisiya [41]3 years ago
6 0

Answer: its 20 x 2

Step-by-step explanation: hope it helps idk if it right but did my best..

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∆ABC has A(-3, 6), B(2, 1), and C(9, 5) as its vertices. The length of side AB is units. The length of side BC is units. The len
leonid [27]

Answer:

AB = 7.07 units

BC = 8.06 units

AC = 12.04 units

Step-by-step explanation:

To find the length for each side of the triangle, apply the distance formula between each pair of vertices.

<u>AB</u>

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(2--3)^2 + (1-6)^2} \\d = \sqrt{(5)^2 + (-5)^2} \\d = \sqrt{25 + 25} \\d = \sqrt{50} \\d=7.07

<u />

<u>BC</u>

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(9-2)^2 + (5-1)^2} \\d = \sqrt{(7)^2 + (4)^2} \\d = \sqrt{49 + 16} \\d = \sqrt{65} \\d=8.06

<u />

<u>AC</u>

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(9--3)^2 + (5-6)^2} \\d = \sqrt{(12)^2 + (-1)^2} \\d = \sqrt{144 + 1} \\d = \sqrt{145} \\d=12.04

4 0
2 years ago
A hyperbola centered at the origin has verticies at (add or subtract square root of 61,0 and foci at (add or subtract square roo
deff fn [24]

Answer:

\frac{x^2}{61}-\frac{y^2}{37}  =1

Step-by-step explanation:

The standard equation of a hyperbola is given by:

\frac{(x-h)^2}{a^2} -\frac{(y-k)^2}{b^2} =1

where (h, k) is the center, the vertex is at (h ± a, k), the foci is at (h ± c, k) and c² = a² + b²

Since the hyperbola is centered at the origin, hence (h, k) = (0, 0)

The vertices is (h ± a, k) = (±√61, 0). Therefore a = √61

The foci is (h ± c, k) = (±√98, 0). Therefore c = √98

Hence:

c² = a² + b²

(√98)² = (√61)² + b²

98 = 61 + b²

b² = 37

b = √37

Hence the equation of the hyperbola is:

\frac{x^2}{61}-\frac{y^2}{37}  =1

6 0
3 years ago
What is 1 7/10 written as an improper fraction
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