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Alex787 [66]
2 years ago
7

Can anyone tell me why by direct substitution of x, the equation (circled ones) equals to the indeterminate form, 0/0? When you

can do x²-k=6 something like that​

Mathematics
1 answer:
Katena32 [7]2 years ago
3 0

Answer:

See explanation and hopefully it answers your question.

Basically because the expression has a hole at x=3.

Step-by-step explanation:

Let h(x)=( x^2-k ) / ( hx-15 )

This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.

Solving for x in that equation:

Adding 15 on both sides:

hx=15

Dividing both sides by h:

x=15/h

For it be a hole, you also must have the numerator is zero at x=15/h.

x^2-k=0 at x=15/h gives:

(15/h)^2-k=0

225/h^2-k=0

k=225/h^2

So if we wanted to evaluate the following limit:

Lim x->15/h ( x^2-k ) / ( hx-15 )

Or

Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.

We were ask to evaluate

Lim x->3 ( x^2-k ) / ( hx-15 )

Comparing the two limits h=5 and k=225/h^2=225/25=9.

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A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What
krok68 [10]

The area of the required triangle is <u>9/π²(3  + √3)</u> sq. units.

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We are asked for the area of the triangle.

Now, the circumference of the circle = 3 + 4 + 5 = 12 units.

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The length of the arcs are proportional to its central angle, making the angles: 3θ, 4θ, and 5θ, which needs to sum up to 360°, giving us θ = 360°/12 = 30°.

Thus, the three arcs subtends angles of θ₁ = 3θ = 90°,θ₂ = 4θ = 120°, and θ₃ = 5θ = 150°.

The area of the circle can be calculated as:

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5 0
1 year ago
Rewrite by completing the square 2x^2+7x+6=0
Kitty [74]

Answer:

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6 0
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