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3 years ago

Solve x 2 - 4x - 7 = 0 by completing the square. What are the solutions?

Mathematics

1 answer:

Bingel [31]3 years ago

3 0

Answer:

x = 2±sqrt(11)

Step-by-step explanation:

x^2 -4x - 7 =0

Add 7 to each side

x^2 -4x -7+7 =0+7

x^2 -4x =7

To complete the square take the coefficient of the x term and divide by 2 and then square it. Add it to both sides

-4 /2 = -2 then square

(-2)^2 = 4

Add 4 to both sides

x^2 -4x+4 =7+4

x^2 -4x+4 = 11

Factor the left side into (x-b/2)^2

(x-2)^2 =11

Take the square root of each side

sqrt((x-2)^2) =±sqrt(11)

x-2 = ±sqrt(11)

Add to to each side

x-2+2 = 2±sqrt(11)

x = 2±sqrt(11)

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pickupchik [31]

This sequence be represented as a recursive equation by a1=8 and an=2a1

<u>Step-by-step explanation</u>:

'Recursive' refers to the repetition of a specific process in a sequence.
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then a2 = 2a(2-1)

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⇒ a2 = 2(8)

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Similarly,

For a2=16,

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3 years ago

A rectangular lamina of uniform density is situated with opposite corners at (0,0) and (15,4). calculate its radii of gyration a

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First, you have to find the moment of inertia along the x and y axes. Constant density is denoted as k.

$I_{x}= \int\limits^15_0\int\limits^4_0 {k y^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

$I_{y}= \int\limits^15_0\int\limits^4_0 {k x^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

Then, the radii of gyration for

x = √[I_x/m]
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where m = k(15-4)² = 121k. Then,

x = y = [4880.33k/121k] = 40.33

I hope I was able to help you. Have a good day.

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Answer:

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