The coordinates of the verticles of

What is the solution to –122 < –3(–2 – 8x) – 8x?

Soloha48 [4]

<span>–122 < –3(–2 – 8x) – 8x
-122< 6 + 24x - 8x
-122 < 6 + 16x
-122 - 6 < 16x
-128<16x
-128/16<x
-8<x
x>-8
answer is </span>B. x > –8

6 0

3 years ago

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PLEASEEEEEE Which system of equations does not have a real solution? y = x 2 + 3 x - 5 and 4 x + 5y = 20 y = x 2 + 3 x - 5 and x

Delicious77 [7]

Answer:

$\Large \boxed{y = x^2 + 3 x - 5 \ \ \mathrm{and} \ \ x + y = -10}$

Step-by-step explanation:

y = x² + 3 x - 5 and x + y = -10 are the system of equations that do not have a real solution.

When the two equations are graphed and if they do not intersect, then there are no real solutions for the system of equations.

8 0

3 years ago

(adapted from Ross, 2.31) Three countries (the Land of Fire, the Land of Wind, and the Land of Earth) each make a 3 person team.

DaniilM [7]

Answer:

The answer is " $\frac{2}{9} \ and \ \frac{1}{9}$ "

Step-by-step explanation:

In point a:

The requires 1 genin, 1 chunin , and 1 jonin to shape a complete team but we all recognize that each nation's team is comprised of 1 genin, 1 chunin, and 1 jonin.

They can now pick 1 genin from a certain matter of national with the value:

$\frac{1}{\binom{3}{1}}=\frac{1}{3} .$

They can pick 1 Chunin form of the matter of national with the value:

$\frac{1}{\binom{3}{1}}=\frac{1}{3} .$

They have the option to pick 1 join from of the country team with such a probability: $\frac{1}{\binom{3}{1}}=\frac{1}{3}$

And we can make the country teams: $3! = 6$ different forms. Its chances of choosing a team full in the process described also are:

$6 \times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{2}{9}.$

In point b:

In this scenario, one of the 3 professional sides can either choose 3 genins or 3 chunines or 3 joniners. So, that we can form three groups that contain the same ninjas (either 3 genin or 3 chunin or 3 jonin).

Its likelihood that even a specific nation team ninja would be chosen is now: $\frac{1}{\binom{3}{1}}=\frac{1}{3}$

Its odds of choosing the same rank ninja in such a different country team are: $\frac{1}{\binom{3}{1}}=\frac{1}{3}$

The likelihood of choosing the same level Ninja from the residual matter of national is: $\frac{1}{\binom{3}{1}}=\frac{1}{3}$ Therefore, all 3 selected ninjas are likely the same grade: $3\times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$