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mr Goodwill [35]
3 years ago
13

Solve for x : 1/a+b+x =1/a+1/b+1/x

Mathematics
1 answer:
ANTONII [103]3 years ago
4 0

Answer:

  x ∈ {-a, -b}

Step-by-step explanation:

  1/(a+b+x) = 1/a +1/b +1/x . . . . given

  abx = bx(a+b+x) +ax(a+b+x) +ab(a+b+x) . . . . multiply by abx(a+b+x)

  (a+b)x^2 +(a+b)^2x +ab(a+b) = 0 . . . . . subtract abx

  x^2 + (a+b)x +ab = 0 . . . . . divide by (a+b)

This is a quadratic equation in x. It will have two solutions, as given by the quadratic formula.

  x = (-(a+b) ±√((a+b)^2 -4(1)(ab))/(2(1)) = (-(a+b) ± |a -b|)/2

Without loss of generality, we can assume a ≥ b, so |a -b| ≥ 0. Then ...

  x = (-a -b -a +b)/2 = -a

  x = (-a -b +a -b)/2 = -b

There are two solutions: x ∈ {-a, -b}.

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Find the coordinates of a point that divides the directed line segment PQ in the ratio 5:3. A) (2, 2) B) (4, 1) C) (–6, 6) D) (4
Yuri [45]

Answer:

The answer is explained below

Step-by-step explanation:

The question is not complete we need point P and point Q.

let us assume P is at (3,1) and Q is at (-2,4)

To find the coordinate of the point that divides a line segment PQ with point P at (x_1,y_1) and point Q at (x_2,y_2) in the proportion a:b, we use the formula:

x-coordinate:\\\frac{a}{a+b}(x_2-x_1)+x_1 \\\\While \ for\ y-coordinate:\\\frac{a}{a+b}(y_2-y_1)+y_1

line segment PQ  is divided in the ratio 5:3 let us assume P is at (3,1) and Q is at (-2,4). Therefore:

x-coordinate:\\\frac{5}{5+3}(-2-3)+3 \\\\While \ for\ y-coordinate:\\\frac{5}{5+3}(4-1)+1

4 0
4 years ago
Ttm question need help
azamat

Answer:

The first option is the correct answer

Step-by-step explanation:

7 0
3 years ago
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3 years ago
Below is an equation being solved.
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Step-by-step explanation:

The math is correct

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Prove this trigonometric equation;<br><br> - tan^2x + sec^2x = 1,
alekssr [168]
Hey there :)

- tan²x + sec²x = 1    or    1 + tan²x = sec²x

sin²x + cos²x = 1 
Divide the whole by cos²x

\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}

\frac{sinx}{cosx} = tanx so \frac{sin^2x}{cos^2x} = tan^2x
and
\frac{1}{cosx} = secx so \frac{1}{cos^2x} = sec^2x

Therefore,
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Take tan²x to the other side {You will have the same answer}

1 = - tan²x = sec²x or sec²x - tanx = 1

3 0
3 years ago
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