How many acute angles are formed by the 4 intersecting lines shown? [asy] import olympiad; size(150); pair a=(0,0), b = 4*dir(0), c=4*dir(40),d=4*dir(100),e=4*dir(15 0),f=4*dir(180),g=4*dir(220),h=4*di r(280),i=4*dir(330); dot(a, red); draw(a--b, endarrow); draw(anglemark(b, a,c,20),black); label("$40^\circ$",anglemark(b, a,c),4*dir(20)); draw(a--c, endarrow); draw(a--d, endarrow); draw(a--e, endarrow); draw(anglemark(d, a,e,20),black); label("$50^\circ$",anglemark(d, a,e),4*dir(120)); draw(a--f, endarrow); draw(a--g, endarrow); draw(a--h, endarrow); draw(anglemark(g, a,h,20),black); label("$60^\circ$",anglemark(g, a,h),4*dir(250)); draw(a--i, endarrow); [/asy]?

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)