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dimulka [17.4K]
3 years ago
12

Find the missing endpoint if S is the midpoint RT. 1. R(9-4) and S(2,-1); fine T

Mathematics
1 answer:
natali 33 [55]3 years ago
8 0

Answer:

The coordinates of point T are (13 , -6)

Step-by-step explanation:

* Lets explain how to solve the problem

- If point (x , y) is the mid point of a line whose end points are

 (x_{1},y_{1}) and (x_{2},y_{2}), then

 x=\frac{x_{1}+x_{2}}{2} and y=\frac{y_{1}+y_{2}}{2}

* Lets solve the problem

∵ Point S is the mid point of segment RT

∴ S = (x , y)

∵ The coordinates of point S are (2 , -1)

∴ x = 2 and y = -1

∵ Point R = (x_{1},y_{1})

∵ The coordinates of point R are (-9 , 4)

∴ x_{1}=-9 and y_{1}=4

∵ Point T = (x_{2},y_{2})

- By using the rule of the mid-point above find the coordinates

 of point T

∴  2=\frac{-9+x_{2}}{2}

- multiply both sides by 2

∴ 4=-9+x_{2}

- Add 9 to both sides

∴ x_{2}=13

∴  -1=\frac{4+y_{2}}{2}

- multiply both sides by 2

∴ -2=4+y_{2}

- Subtract 4 from both sides

∴ y_{2}=-6

∴ The coordinates of point T are (13 , -6)

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A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi
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Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

= \frac{20!}{(20-5)! 5!)}

= \frac{20*19*18*17*16}{5*4*3*2*1}

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

= \frac{20}{320} = \frac{1}{19}

P(A5) = ⁸C₅ / ²⁰C₅

= \frac{56}{15540} = \frac{7}{1938}

P(A4) = ⁷C₄ / ²⁰C₄

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c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0
3 years ago
Fill in the blank with a constant, so that the resulting quadratic expression is the square of a binomial. $\[x^2 + 22x + \under
saw5 [17]

Answer:

$\[x^2 + 22x + 121\]$

Step-by-step explanation:

Given

$\[x^2 + 22x + \underline{~~~~}.\]$

Required

Fill in the gap

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$\[x^2 + 22x + k\]$

Solving for k...

To do this, we start by getting the coefficient of x

Coefficient of x = 22

<em />

Divide the coefficient by 2

Result = 22/2

Result = 11

Take the square of this result, to give k

k= 11^2

k= 121

Substitute 121 for k

$\[x^2 + 22x + 121\]$

The expression can be factorized as follows;

x^2 + 11x + 11x + 121

x(x + 11)+11(x+11)

(x+11)(x+11)

(x+11)^2

<em>Hence, the quadratic expression is </em>$\[x^2 + 22x + 121\]$<em></em>

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mixas84 [53]

Answer:

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tan θ = 28.2/45.8

θ = arc tan (28.2/45.8)

θ = 31.621459805°

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