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3 years ago

What are the roots of the quadratic equation below? x 2 + 2x = -5

2 answers:

6 0

To find the roots of the quadratic equation x^2 + 2x + 5 = 0 is the same as solving it for x.
The formula to get x₁ and x₂ is: x₁,x₂=(-b⁺/₋√(b²-4ac))/(2a) where in our case a=1, b=2 and c=5.

Lets input the numbers:

x₁,x₂= (-2⁺/₋√(2²-4*1*5))/(2*1) = (-2⁺/₋√(4-20))/2, = (-2⁺/₋√(-16))/2

We see that we have a minus sign under the square root so the solutions or roots for our quadratic equation are going to be complex numbers:

x₁ = (-2+4i)/2 = -1+2i

x₂ = (-2-4i)/2 = -1-2i

So our roots are complex and are: x₁= -1+2i and x₂= -1-2i.

steposvetlana [31]3 years ago

3 0

This is the concept of quadratic equations, to get the root of quadratic equation given we proceed as follows;
x^2+2x=-5
Writing the above in the quadratic form ax^2+bx+c=0 we get;
x^2+2x+5=0
(x+1)^2+4=0
x=-1-2i
x=-1+2i
where i=√-1

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5. GEOMETRY The volume of a rectangular pyramid is one third the product of the area of its base and its height. Find

tatuchka [14]

Answer:

x³ + 7x² + 15x + 9

Step-by-step explanation:

Given that the volume (V) of a rectangular pyramid is

V = $\frac{1}{3}$ A h ( A is the area of base and h the height ), then

V = $\frac{1}{3}$ (3x² + 12x + 9)(x + 3) ← factor out 3 from A

= $\frac{1}{3}$ × 3(x² + 4x + 3)(x + 3)

= (x² + 4x + 3)(x + 3)

Each term in the second factor is multiplied by each term in the first factor, that is

x² (x + 3) + 4x(x + 3) + 3(x + 3) ← distribute parenthesis

= x³ + 3x² + 4x² + 12x + 3x + 9 ← collect like terms

= x³ + 7x² + 15x + 9

Thus the expression for the volume is

V = x³ + 7x² + 15x + 9

4 0

3 years ago

Two right triangles are graphed on a coordinate plane. One triangle has a vertical side of 2 and a horizontal side of 6. The oth

Vikki [24]

Consider two right triangles:

1. ΔABC with vertices A(0,0), B(0,2), C(6,0). Then AB is perpendicular to AC and AB=2 units (vertical leg), AC=6 units (horizontal leg).

2. ΔXYZ with vertices X(6,-10), Y(6,0), Z(36,-10). Then XY is perpendicular to XZ and XY=10 units (vrrtical leg), XZ=30 units (horizontal leg).

The equation of the line BC is

$\dfrac{x-0}{6-0}=\dfrac{y-2}{0-2},\\ \\-x=3y-6,\\ \\x+3y-6=0.$

Check whether points Y and Z lie on this line:

Y(6,0): $6+3\cdot 0-6=0$ - true;

Z(36,-10): $36+3\cdot (-10)-6=0$ - true.

Answer: the hypotenuses of these two triangles could lie along the same line

3 0

3 years ago

If one point on a horizontal line has coordinates (3, 2), which of the points below are also on the line? Select all that apply.

stepan [7]

Answer:

If the line is horizontal, then all that have a 2 in the Y space will apply because the line goes across meaning that it won't change on the vertical axis.

(5,2) and (-2, 2)

6 0

4 years ago

It is estimated that t-weeks into a semester, the average amount of sleep a college math student gets per day S(t) decreases at

goldenfox [79]

Answer:

s(15)= $\frac{-3}{2} e^{15^{2} } +9.6$

Step-by-step explanation:

Before to getting started ,we have to consider the function which represents sleep decline rate S'(t) = $-3te^{t^{2} } \\$

Since this function is interpreted in this case as a rate of change, we can deduce this funcion is the first derivative of the function of average amount of sleep (S(t)).

In order to find S(t), we will integrate S'(t)

$\int\limits^ {} -3t3e^{t^{2} } \, dx \\$

This case corresponds to intergration by substitution

Substituting u= $x^{2} \\$

Computting the differential of u

$\frac{du}{dt} =2tdt\\$

Now, let's go back to our integral and write it in terms of u, using both definition of u and its defferential.

$\int\limits^{}-3te^{u} \, \frac{1}{2t}du \\\\\int\limits^{}\frac{-3}{2} e^{u}du\\$

The integral of a exponential function is itself. So the result is:

$\frac{-3}{2} e^{u} +c$

Now substitute u back in:

$\frac{-3}{2} e^{t^{2} } +c$

This answer correspond to function of average amount of sleep.

We know that when the semester begins (t=0) , math students sleep an average of 8.1 hours per day so we will use this info in order to find the value of c.

$s(0)=\frac{-3}{2} e^{0^{2} } +c=8.1\\\\\frac{-3}{2} +c=8.1$