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stellarik [79]
3 years ago
13

show that thw roots of the equation (x-a)(x_b)=k^2 are always real if a,b and k are real. Please I really need help with this

Mathematics
1 answer:
VLD [36.1K]3 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

Check the value of the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real

• If b² - 4ac = 0 roots are real and equal

• If b² - 4ac < 0 then roots are not real

given (x - a)(x - b) = k² ( expand factors )

x² - bx - ax - k² = 0 ( in standard form )

x² + x(- a - b) - k² = 0

with a = 1, b = (- a - b), c = -k²

b² - 4ac = (- a - b)² + 4k²

For a, b, k ∈ R then (- a - b)² ≥ 0 and 4k² ≥ 0

Hence roots of the equation are always real for a, b, k ∈ R


           

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Please add explanation as well.
Marianna [84]
It is the first one. Three times a number means 3x. The sum of 3x and 7 is 3x + 7 because sum means additions. And 3x + 7 is greater than four times the number which is 4x and greater than is this symbol >. So 3x + 7 > 4x
3 0
3 years ago
If point A is located at (-7,-3) and there are 12 points between A and B what could be the possible point coordinates for point
Ivahew [28]
(5,-3), (-7,9) are two possible point coordinates
6 0
3 years ago
Plzzzzz answer this question about density mass and volume!!!!!!!!!
nataly862011 [7]

Answer:

13.896 kg

Step-by-step explanation:

You can find the mass of the bar by first finding the volume.

V = BH

where B = area of the base (the trapezium), and

H = height (distance trapezium between bases)

The area of a trapezium is

A = (b1 + b2)h/2

where b1 and b2 are the lengths of the bases of the trapezium (the parallel sides), and

h = the altitude of the trapezium (distance between the bases of the trapezium)

V = (b1 + b2)h/2 * H

V = (12 cm + 6 cm)(5 cm)/2 * 16 cm

V = 720 cm^3

The volume of the bar is 720 cm^3.

Now we use the density and the volume to find the mass.

density = mass/volume

mass = density * volume

mass = 19.3 g/cm^3 * 720 cm^3

mass = 13,896 g

Now we convert grams into kilograms.

1 kg = 1000 g

mass = 13,896 g * (1 kg)/(1000 g)

mass = 13.896 kg

Answer: 1.3896 kg

6 0
3 years ago
Lim t^4 - 6 / 2t^2 - 3t + 7
Harman [31]

I think you meant to say

\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}

(as opposed to <em>x</em> approaching 2)

Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}

Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2:

\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}

6 0
3 years ago
What does two and five fiths equal to
erma4kov [3.2K]

2 and five fifths = 3


I hope that's help. Good night .

6 0
3 years ago
Read 2 more answers
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