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lions [1.4K]
3 years ago
13

A. 5y2 - 60 = 0by Square root property​

Mathematics
1 answer:
TEA [102]3 years ago
7 0

Answer:

y  = ± 2sqrt(3)

Step-by-step explanation:

5y^2 - 60 = 0

Add 60 to each side

5y^2  = 60

Divide by 5

y^2 = 60/5

y^2 = 12

Take the square root of each side

sqrt( y^2) = ± sqrt(12)

y = ± sqrt(4*3)

y =  ±sqrt(4) sqrt(3)

y  = ± 2sqrt(3)

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2 years ago
Read 2 more answers
An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh
natima [27]

Answer:

Speed of the plane in still air: 779\; {\rm km \cdot h^{-1}}.

Windspeed: 100\; {\rm km \cdot h^{-1}}.

Step-by-step explanation:

Assume that x\; {\rm km \cdot h^{-1}} is the speed of the plane in still air, and that y\; {\rm km \cdot h^{-1}} is the speed of the wind.

  • When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be (x - y)\; {\rm km \cdot h^{-1}}.
  • When this plane is travelling in the same direction as the wind, the ground speed of this plane would be (x + y)\; {\rm km \cdot h^{-1}}.

The question states that when going against the wind (v = (x - y)\; {\rm km \cdot h^{-1}},) the plane travels 6111\; {\rm km} in 9\; {\rm h}. Hence, 9\, (x - y) = 6111.

Similarly, since the plane travels 7911\; {\rm km} in 9\; {\rm h} when travelling in the same direction as the wind (v = (x + y)\; {\rm km \cdot h^{-1}},) 9\, (x + y) = 7911.

Add the two equations to eliminate y. Subtract the second equation from the first to eliminate x. Solve this system of equations for x and y: x = 779 and y = 100.

Hence, the speed of this plane in still air would be 779\; {\rm km \cdot h^{-1}}, whereas the speed of the wind would be 100\; {\rm km \cdot h^{-1}}.

3 0
1 year ago
The vertex of this parabola is at (-3,2). Which of the following could be its equation?
VashaNatasha [74]

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7 0
2 years ago
Read 2 more answers
Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i
aleksandrvk [35]

Answer:

0.967 = 96.7% probability the rock sample actually contains raritanium

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive reading

Event B: Contains raritanium

Probability of a positive reading:

98% of 13%(positive when there is raritanium).

0.5% of 100-13 = 87%(false positive, positive when there is no raritanium). So

P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175

Positive when there is raritanium:

98% of 13%

P(A) = 0.98*0.13 = 0.1274

What is the probability the rock sample actually contains raritanium?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967

0.967 = 96.7% probability the rock sample actually contains raritanium

7 0
2 years ago
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