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xenn [34]
3 years ago
11

The co-ordinate of point P is (x+3) and Q is (x-3). Then find d(P,Q). If coordinate of R is (x+2). Find d(Q,R).

Mathematics
1 answer:
kari74 [83]3 years ago
5 0

Answer:

<h2>d(P, Q) = 6</h2><h2>d(Q, R) = 5</h2>

Step-by-step explanation:

The formula of a distance between two points A(a) and B(b):

<h3>d = |b - a|</h3>

We have P(x + 3), Q(x- 3), R(x + 2).

Substitute:

d(P, Q) = |(x - 3) - (x + 3)| = |x - 3 - x - 3| = |-6| = 6

d(Q, R) = |(x + 2) - (x - 3)| = |x + 2 - x - (-3)| = |2 + 3| = |5| = 5

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wel

Answer:

A and A is the answer. so this is the answer

7 0
3 years ago
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If f(x) = x-6 and g(x)= 1/2x (x+3), find g(x) * f(x)
sertanlavr [38]

Answer:

Final answer is g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}.

Step-by-step explanation:

given functions are f(x)=x-6 and g\left(x\right)=\frac{1}{2x\left(x+3\right)}.

Now we need to find about what is the value of g\left(x\right)*f\left(x\right).

g\left(x\right)*f\left(x\right) simply means we need to multiply the value of  f(x)=x-6 and g\left(x\right)=\frac{1}{2x\left(x+3\right)}.

g\left(x\right)\cdot f\left(x\right)=\frac{1}{2x\left(x+3\right)}\cdot\left(x-6\right)

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Hence final answer is g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}.

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3 years ago
Does anyone know how to solve this?
Bezzdna [24]

Use this version of the Law of Cosines to find side b:

b^2 = a^2 + c^2 − 2ac cos(B)

We want side b.

b^2 = (41)^2 + (20)^2 - 2(41)(20)cos(36°)

After finding b, you can use the Law of Sines to find angles A and C or use other forms of the Law of Cosines to find angles A and C.

Try it....

6 0
4 years ago
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Answer:

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Harlamova29_29 [7]

Answer:

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Interval Notation: (-∞, -6)

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