If the average (arithmetic mean) of 3a and 4b is less than 50, and a is twice b, what is the largest integer value of a?

The largest value of 'a' is less than 20

Step-by-step explanation:

If arithmetic mean of 3a and 4b is less than 50 i.e 3a+4b/2 < 50

3a+4b < 100... (1)

If a is twice as b, a = 2b... (2)

Substituting equation 2 into 1 we have, 3(2b)+4b < 100

6b+4b < 100

10b < 100

b < 10

Since a = 2b, a = 2×10 = 20

Therefore the largest value of 'a' is less than 20 which can be 19.9

22

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3