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3 years ago

Rectangle 2. triangle 3. cone 4. circle 5. cylinder

Mathematics

1 answer:

7nadin3 [17]3 years ago

6 0

Circle = perpendicular cross section of a cylinder

cone = perpendicular cross section of a pyramid

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For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

andriy [413]

Answer:

If $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

The signal of $\bigtriangleup$ determines how many real roots an equation has:

$\bigtriangleup > 0$ : Two real and different solutions

$\bigtriangleup = 0$ : One real solution

$\bigtriangleup < 0$ : No real solutions

In this problem, we have the following second order polynomial:

$(k+1)x^{2} + 4kx + 2 = 0$ .

This means that $a = k+1; b = 4k; c = 2$

It has one solution if

$\bigtriangleup = 0$

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We can simplify by 8

$2k^{2} - k - 1 = 0$

The solution is:

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So, if $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

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3 years ago

Can someone help me please?

lakkis [162]

Choice A is correct
BC=DC
the two triangles share one angel C
the two triangles each have a right angle
so they are congruent according to Angle-Side_Side Theorem

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