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3 years ago

Explain the difference between quadratic equations with one solution, two solutions, and complex solutions

1 answer:

4 0

Recall that given the equation of the second degree (or quadratic)
ax ^ 2 + bx + c
Its solutions are:
x = (- b +/- root (b ^ 2-4ac)) / 2a
discriminating:
d = root (b ^ 2-4ac)
If d> 0, then the two roots are real (the radicand of the formula is positive).
If d = 0, then the root of the formula is 0 and, therefore, there is only one solution that is real and of multiplicity 2 (it is a double root).
If d <0, then the two roots are complex and, in addition, one is the conjugate of the other. That is, if one solution is x1 = a + bi, then the other solution is x2 = a-bi (we are assuming that a, b, c are real).
One solution:
A cut point with the x axis
Two solutions:
Two cutting points with the x axis.
Complex solutions:
Does not cut to the x axis

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Evaluate the surface integral ∫sf⋅ ds where f=⟨2x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with

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Parameterize S by the vector function

$\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

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The integral of the field over S is then

$\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \vec f(\vec s) \cdot \vec n \, du \, dv$

$\displaystyle = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv$

$\displaystyle = 128 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \cos^2(u) \sin^3(v) \, du \, dv = \boxed{\frac{64\pi}3}$

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2 years ago

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3 years ago

An isosceles trapezoid has base angles of 45° and bases of lengths 8 and 14. The area of the trapezoid is

Mice21 [21]

Answer:

Option A is correct.

Step-by-step explanation:

Given an isosceles trapezoid has base angles of 45° and bases of lengths 8 and 14. we have to find the area of isosceles trapezoid.

An isosceles trapezoid has base angles of 45° and bases of lengths 8 and 14.

From the figure attached , we can see an isosceles trapezoid ABCD,

AB = 8cm and CD=14cm

So we have to find the value of AE which is the height of Trapezoid in order to find area.

In ΔAED

$tan\angle 45 =\frac{AE}{ED}$

⇒ $AE=1\times 3$

∴ AE = DE =3cm

$\text{The area of the trapezoid=}\frac{h}{2}\times (a+b)$

h=3cm, a=14cm, b=8cm

$Area=\frac{3}{2}\times(14+8)=\frac{3}{2}\times 22=33 units^2$

hence, $\text{The area of the trapezoid is }33 units^2$

Option A is correct.

3 0

3 years ago

Read 2 more answers

Help! giving 13 points!

Shtirlitz [24]

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