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marysya [2.9K]
3 years ago
7

URGENT What is the distance between the two points (-5, 6) and (8,-4)?

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
8 0

Given parameters:

Coordinates of the two points = (-5, 6) and (8,-4)

Unknown:

Distance between the two points.

To find the distance between two points, use the expression below;

  Distance ²  = (x₂ - x₁)² + (y₂ - y₁)²

where;

(-5, 6) and (8,-4)

         x₂ = 8 , x₁ = -5 , y₂ = -4 and y₁ = 6

Input the parameters and solve;

 Distance²  =  (8-(-5))²  +  (-4 - (6))²

                    = 169 + 100

                    = 269

Distance = \sqrt269}   = 16.4

The distance between the two points is 16.4

Kaylis [27]3 years ago
4 0

Answer:

x1 = -5   x2 = 8

y1 = 6   y2 = -4

Distance = Square Root ((x2 - x1) ^2 + (y2 -  y1)^2)=

Sqr root [(8 --5) ^ 2 + (-4 -6) ^ 2] =

Sqr root [13^2 + (-10)^2] =

Sqr root [169 + 100]

= Sqr root [269] = 16.4012194669

Source: http://www.1728.org/distance.htm

Step-by-step explanation:

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VashaNatasha [74]
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5 + 6 + 5 = 16

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5:16

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3 years ago
Assume that you assign the following subjective probabilities for your final grade in your econometrics course (the standard GPA
rodikova [14]

Answer:

E(X) = 4*0.2 +3*0.5+ 2*0.2 +1*0.8+ 0*0.02= 2.78

So then the best answer for this case would be:

C. 2.78

Step-by-step explanation:

For this case we have the following probabability distribution function given:

Score         P(X)

A= 4.0       0.2

B= 3.0       0.5

C= 2.0       0.2

D= 1.0        0.08

F= 0.0       0.02

______________

Total          1.00

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

If we use the definition of expected value given by:

E(X) = \sum_{i=1}^n X_i P(X_I)

And if we replace the values that we have we got:

E(X) = 4*0.2 +3*0.5+ 2*0.2 +1*0.8+ 0*0.02= 2.78

So then the best answer for this case would be:

C. 2.78

3 0
3 years ago
PLS HELP I’ve been trying to do this but I just can’t do it
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Answer:

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Step-by-step explanation:

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3 years ago
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Klio2033 [76]

Callum must buy one packet of dog food every 10 days

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2 years ago
Perform a first derivative test on the function ​f(x)equals2 x cubed plus 3 x squared minus 120 x plus 6​; ​[minus5​,8​]. Bold
maria [59]

Answer:

a) Critical points

x = 4 and x = -5

b) x = 4 corresponds to a minimum point for the function f(x)

x = - 5 corresponds to a maximum point for the function f(x)

c) The minimum value of f(x) in the interval = -298

The maximum value of f(x) in the interval = 431;

Step-by-step explanation:

f(x) = 2x³ + 3x² - 120x + 6 in the interval [-5, 8]

a) To obtain the critical points, we need to obtain the first derivative of the function with respect to x. Because at critical points of a function, f(x), (df/dx) = 0

f'(x) = (df/dx) = 6x² + 6x - 120 = 0

6x² + 6x - 120 = 0

Solving the quadratic equation,

x = 4 or x = -5

The two critical points, x = 4 and x = -5 are in the interval given [-5, 8] (the interval includes -5 and 8, because it's a closed interval)

b) To investigate the nature of the critical points, we obtain f"(x)

Because f'(x) changes sign at the critical points

If f"(x) > 0, then it's a minimum point and if f"(x) < 0 at c, then it's a maximum point.

f"(x) = (d²f/dx²) = 12x + 6

at critical point x = 4

f"(x) = 12x + 6 = 12(4) + 6 = 54 > 0, hence, x = 4 corresponds to a minimum point.

at critical point x = -5

f"(x) = 12x + 4 = 12(-5) + 4 = -56 < 0, hence, x = -5 corresponds to a maximum point.

c) At x = 4,

f(x) = 2x³ + 3x² - 120x + 6 = 2(4)³ + 3(4)² - 120(4) + 6 = -298

At x = - 5

f(x) = 2x³ + 3x² - 120x + 6 = 2(-5)³ + 3(-5)² - 120(-5) + 6 = 431

7 0
3 years ago
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