There is a point $A$ with positive coordinates such that the sum of the coordinates of $A$ is $14$. If the $x$-coordinate of $A$ is $a$, then point $P$ is at $(3a, a^2+13a-11)$. If the slope of a line passing through $A$ and $P$ is $7$, find $a$.

  5

Step-by-step explanation:

Given:

  A = (a, 14-a)

  P = (3a, a^2 +13a -11)

  the slope of AP is 7

  a > 0

Find:

  a

Solution:

The slope of AP is ...

  m = (Py -Ay)/(Px -Ax)

  7 = (a^2 +13a -11 -(14 -a))/(3a -a)

  14a = a^2 +14a -25

  25 = a^2

  a = √25 = 5 . . . . . the positive solution

The value of 'a' is 5.

Check

The point A is (a, 14-a) = (5, 9).

The point P is (3a, a^2 +13a -11) = (15, 79)

The slope of AP is (79 -9)/(15 -5) = 70/10 = 7.

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier