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zloy xaker [14]
2 years ago
3

In the diagram, PR and AC are vertical lines, while CB is a horizontal line segment. If CR : RB = 2 : 3, what are the coordinate

s of point Q?. A.(-2, -4) B.(-1, -5) C.(-3, -3) D.(0, -6)
Mathematics
2 answers:
denpristay [2]2 years ago
4 0
We are given that the ratio of CR to RB is 2:3. We can apply similar triangles to determine the coordinates of Point q. Using similar triangle theorem, the formula that applies is BR / BC = BQ/ BABA has a length of 5 sqrt 2. BR/BC is equal to 3/5. substituing, the formula becomes 3 / 5 = BQ/ 5 sqrt 2BQ = 3 sqrt 2. the point to give a distance that of BQ is A (-2,-4)
zhenek [66]2 years ago
3 0

Answer:

The correct answer is

A (-2,-4)

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Milton spilled some ink on his homework paper. He can't read the coefficient of $x$, but he knows that the equation has two dist
Lera25 [3.4K]

Answer:

Sum = -81

Step-by-step explanation:

See the comment for complete question.

Given

c = 36 ----- Constant

No coefficient of x^2

Required:

Determine the sum of all distinct positive integers of the coefficient of x

Reading through the complete question, we can see that the question has 3 terms which are:

x^2 ---- with no coefficient

x ---- with an unknown coefficient

36 ---- constant

So, the equation can be represented as:

x^2 + ax + 36 = 0

Where a is the unknown coefficient

From the question, we understand that the equation has two negative integer solution. This can be represented as:

x = -\alpha and x = -\beta

Using the above roots, the equation can be represented as:

(x + \alpha)(x + \beta) = 0

Open brackets

x^2 + (\alpha + \beta)x + \alpha \beta = 0

To compare the above equation to x^2 + ax + 36 = 0, we have:

a = \alpha + \beta

\alpha \beta = 36

Where: \alpha, \beta and \alpha \ne \beta

The values of \alpha and \beta that satisfy \alpha \beta = 36 are:

\alpha = -1 and \beta = -36

\alpha = -2 and \beta = -18

\alpha = -3 and \beta = -12

\alpha = -4 and \beta = -9

So, the possible values of a are:

a = \alpha + \beta

When \alpha = -1 and \beta = -36

a = -1 - 36 = -37

When \alpha = -2 and \beta = -18

a = -2 - 18 = -20

When \alpha = -3 and \beta = -12

a = -3 - 12 = -15

When \alpha = -4 and \beta = -9

a = -4 - 9 = -13

At this point, we have established that the possible values of a are: -37, -20, -15 and -9.

The required sum is:

Sum = -37 -20 -15 - 9

Sum = -81

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Answer:

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Answer:

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