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3 years ago

Which of the following equations have complex roots?

Mathematics

1 answer:

valkas [14]3 years ago

8 0

Answer:

Step-by-step explanation:

Completing the square: (x+ 7/2)^2 = -2 + 49/4 = 41/4, in the next step you'll be square rooting a positive number so it'll have real roots.

Completing the square: (x+3/2)^2 = -9 + 9/4 = -27/4, you'll next be square rooting a negative number, so it'll have complex roots

Completing the square: (x-5/2)^2 = -1 + 25/4 = 21/4, next you'll square root a positive number, so real roots

Completing the square: (x+7/2)^2 = 2 + 49/4 = 57/4, square root a positive number, so real roots.

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Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

You're minimizing $d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}$ subject to the constraint $f(x,y,z)=x+y+z=1$ . Note that $d(x,y,z)$ and $d(x,y,z)^2$ attain their extrema at the same values of $x,y,z$ , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .

7 0

3 years ago

WILL GIVE BRAINLIEST!

andreev551 [17]

Answer:

1,632(

Step-by-step explanation:

I did 34,284/252 to figure out the # of groups with 252 people

I got 136.04-- rounded it to 136

I then multiplied 136*12=1,632

so 1,632 people have type A blood

3 0

3 years ago

Read 2 more answers

Here is Laila’s graph from the previous screen.

lidiya [134]

Answer:

y=-1/3x+8

Step-by-step explanation:

There is no need for any specific answers, but here is one that could logically work out. Since the graph is going left/down, it has a negative slope, so -1/3 would be reasonable. The graph doesn't cross the origin and crosses above it, so this equation must have a positive 'b' value. In this case, I chose 8. y=-1/3+8 could represent Laila's graph.

4 0

2 years ago

in a five character password the first two characters must be digits and the last three characters must be letters if no charact

Vilka [71]

Answer:

1,404,000 unique passwords are possible.

Step-by-step explanation:

The order in which the letters and the digits are is important(AB is a different password than BA), which means that the permutations formula is used to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$