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Irina18 [472]
3 years ago
13

If AACD ~ AABE, find the value of x.

Mathematics
1 answer:
Anastaziya [24]3 years ago
4 0

9514 1404 393

Answer:

  x = 16

Step-by-step explanation:

Corresponding sides are proportional.

  BE/CD = AE/AD

  20/(3x+8) = (x-1)/(x-1+27)

  20(x+26) = (3x+8)(x -1)

In standard form, this is ...

  3x^2 -15x -528 = 0

  x^2 -5x -176 = 0 . . . . . . divide by 3

  (x -16)(x +11) = 0 . . . . . . . factor

  x = 16

__

This is the positive value that makes the product zero. x=-11 will also make the product 0, but gives negative segment lengths in the geometry. It is an extraneous solution.

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Geometric Sequence:  

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Step-by-step explanation:


8 0
3 years ago
A 6 cm long spring extends to 9 cm when a 1 kg load is suspended from it. What would be its length if a 2 kg load were suspended
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12 cm

Step-by-step explanation:

To calculate the length of a spring with a 2 kg load, compare the displacement of a 1 kg load and adjust accordingly.

When a 1 kg load is suspended from the spring, the spring which is 6 cm stretches to 9 cm. This is 3 cm longer due to the weight. If you attach a weight which is twice as much then the displacement will be twice as much. Instead of stretching an additional 3 cm, it will stretch 2*3 = 6 cm. Add this to the length of the spring and it stretches in total 6 + 6 = 12 cm.

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8 0
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Read 2 more answers
Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0
natita [175]

Answer:

\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C

Step-by-step explanation:

Given differential equation,

(x^2 + y^2) dx + (x^2 - xy) dy = 0

\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)

Let y = vx

Differentiating with respect to x,

\frac{dy}{dx}=v+x\frac{dv}{dx}

From equation (1),

v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}

v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}

v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}

v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}

x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v

x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}

x\frac{dv}{dx}=\frac{v+1}{v-1}

\frac{v-1}{v+1}dv=\frac{1}{x}dx

Integrating both sides,

\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx

\int{\frac{v-1+1-1}{v+1}}dv=lnx + C

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Now, y = vx ⇒ v = y/x

\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C

5 0
3 years ago
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