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love history [14]
3 years ago
10

____ two points are collinear

Mathematics
2 answers:
kaheart [24]3 years ago
8 0
Any... That would be the answer
dmitriy555 [2]3 years ago
5 0
Two points are trivially collinear since two points determine a line.
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mark loads a crate that weighs 22.5 pounds and 12 boxes that each weigh 11.25 pounds onto a truck. What is the total weight of t
gladu [14]
The total weight is 157.5 pounds. 11.25 x 12 is 135. 135 plus 22.5 is 157.5
7 0
3 years ago
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Find the x value for the point that divides the lines segment AB into ratio of 2:3
m_a_m_a [10]

Let (x_A,y_A),\ (x_B,y_B) be the coordinates of the points A and B and (x_0,y_0) be the coordinates of the point O that divides the segment AB in ratio 2:3.

Consider vectors

\overrightarrow{AO}=(x_0-x_A,y_0-y_A),\\ \\\overrightarrow{OB}=(x_B-x_0,y_B-y_0).

These vectors are collinear and

\dfrac{\overrightarrow{AO}}{\overrightarrow{OB}}=\dfrac{2}{3}.

Then

\left\{\begin{array}{l}\dfrac{x_0-x_A}{x_B-x_0}=\dfrac{2}{3}\\ \\\dfrac{y_0-y_A}{y_B-y_0}=\dfrac{2}{3}\end{array}\right.\Rightarrow \left\{\begin{array}{l}3(x_0-x_A)=2(x_B-x_0)\\ \\3(y_0-y_A)=2(y_B-y_0)\end{array}\right..

This means that

\left\{\begin{array}{l}3x_0-3x_A=2x_B-2x_0\\ \\3y_0-3y_A=2y_B-2y_0\end{array}\right.\Rightarrow \left\{\begin{array}{l}5x_0=2x_B+3x_A\\ \\5y_0=2y_B+3y_A\end{array}\right..

Thus,

x_0=\dfrac{2x_B+3x_A}{5},\ y_0=\dfrac{2y_B+3y_A}{5}.

Answer: \left(\dfrac{2x_B+3x_A}{5},\dfrac{2y_B+3y_A}{5}\right).

8 0
4 years ago
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Conditional Probability
Anna35 [415]
For this problem take the amount of people who ordered large hot drinks (22) and divide it by the total amount of people that ordered hot drinks (75)

And the answer is .293333333 or rounded to 29%
8 0
3 years ago
How many strings of eight uppercase English letters are there if no letter can be repeated?
Gemiola [76]
............................3................................
7 0
4 years ago
Which function can be used to model the graphed geometric sequence?
andreev551 [17]

Answer: The correct option is 1, i.e., f(x+1)=\frac{5}{6}f(x).

Explanation:

The geometric sequence is in the form of,

a,ar,ar^2,ar^3,....

Where, a is the first term of the sequence and r is the common ratio of the sequence.

It means the n_{th} term of the sequence is defined as,

a_n=ar^{n-1}

So the the (n+1)_{th} term of the sequence is defined as,

a_{n+1}=ar^n

a_{n+1}=r(ar^{n-1})

a_{n+1}=ra_n

It means the geometric sequence is in the form of,

f(x+1)=rf(x)

Where, r be any constant.

From the options only f(x+1)=\frac{5}{6}f(x) is in the form of f(x+1)=rf(x) with common ratio \frac{5}{6}.

Therefore, the function can be used to model the graphed geometric sequence is f(x+1)=\frac{5}{6}f(x) .

8 0
4 years ago
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