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3 years ago

Find the x and y intercepts, show all work. 6x+3y=12

Mathematics

1 answer:

vampirchik [111]3 years ago

8 0

Y=-2x+4
you can plug any point for x and get the y based on the x point

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On a fruit farm 1/3 of the trees are apple trees and 1/5 are lemon trees. The rest are fig trees. There are 90 trees in all.

Vaselesa [24]

Answer:

I'm not sure what the question was but I've worked out all the three types of trees

Step-by-step explanation:

⅓ + ⅕ = 8/15 because:

×5 ×3

5/15 + 3/15 = 8/15

therefore 7/15 is fig trees

90÷15 =6

6×5 = 30 apple trees

6×3 = 18 lemon trees

6×7 = 42 fig trees

7 0

3 years ago

1/3 of your birthday cake is leftover from your party. If you eat 1/4 of the leftover cake, what fraction of the original birthd

navik [9.2K]

Answer:

a twelth= 1/12

Step-by-step explanation:

3 0

3 years ago

Due very soon so if someone could answer soon-ish it would be very helpful and much appreciated <3<3<3

scoundrel [369]

The answer is 1 i did this before and hope this helps

5 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: $3\times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$