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2 years ago

How to find the midpoint using coordinates A(-1,5), B(2,-3)

2 answers:

5 0

The midpoint formula is (x1+x2)/(2),(y1+y2)/(2). The coordinates will be substituted into this formula: (-1+2)/(2), (5+-3)/(2). This simplifies to (1/2,1). This is the midpoint. Hope this helped!

Alex Ar [27]2 years ago

4 0

Count the line and the distance between them

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Please answer this question now

Alex787 [66]

Answer:

541.7 (m2)

Step-by-step explanation:

Applying the sine theorem:

WV/sin(X) = XV/sin(W)

=> WV = XV*sin(X)/sin(W) = 37*sin(50)/sin(63) = 31.81

Angle V = 180 - X - W = 180 - 50 - 63 = 67

Denote WH is a height of the triangle VWX, H lies on XV

=> WH = WV*sin(V) = 31.81*sin(67) = 29.28

=> Area of triangle VWX is calculated by:

S = side*height/2 = XV*WH/2 = 37*29.28/2 = 541.7 (m2)

8 0

2 years ago

Write an equation to represent each situation.

mina [271]

Answer:

18 years

Step-by-step explanation:

2(8+x) = 34+x

16+2x = 34+x

-18 = -x

4 0

2 years ago

Given: m∠ABC = m∠CBD Prove: Ray B C bisects ∠ABD. 3 lines extend from point B. One line extends to point A, another to C, and an

insens350 [35]

Answer:

a: given

b: definition of congruent

c: definition of bisect

Step-by-step explanation:

just took the assignment

8 0

2 years ago

Read 2 more answers

A paper container in the shape of a right circular cone has a radius of 3 inches and a height of 8 inches. Determine and state t

Scrat [10]

Given:

Radius of the cone = 3 inches

Height of the cone = 8 inches

To find:

The volume of the cone, in terms of π.

Solution:

We know that, the volume of a cone is:

$V=\dfrac{1}{3}\pi r^2h$

Putting $r=3,h=8$ , we get

$V=\dfrac{1}{3}\pi (3)^2(8)$

$V=\dfrac{1}{3}\pi (9)(8)$

$V=\dfrac{72}{3}\pi$

$V=24\pi$

Therefore, the volume of the cone is $24\pi$ .

7 0

2 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) Probability of distance at most 100m, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

Probability of distance at most 200:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

Probability of distance between 100 and 200:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

Distance exceeds the mean distance by more than 2σ:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

3 years ago