A log is 20 m long, correct to the nearest metre. It has to be cut into fence posts which must be 90 cm long, correct
to the nearest 10 centimetres.
What is the largest number of fence posts that can possibly be cut from the log?

222 fence posts

Step-by-step explanation:

Since the log is 20 m long, correct to the nearest metre and has to be cut into fence posts which must be 90 cm long, correct  to the nearest 10 centimetres, there can be x fence posts 90 cm long.

So, we have x × 90 cm = 20 m

x × 0.9 m = 20 m

x = 20 m/0.9 m

x = 222.22

x ≅ 222 fence posts

So, the largest number of fence posts that can possibly be cut from the log is 222 fence posts.

-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

5x + 5

step-by-step explanation:

you should always add the diagrams and graphs we cannot see. this particular question is kind of borderline because we can make a graph for you.

(f + g)x = -x^2 +3x + 5 + x^2 + 2x

(f + g)x = 5x + 5

the graph of z = 5x + 5 is shown below. always use desmos when in doubt.