we have
(a+b)(3a-b)(2a+7b)
step 1
Solve
apply distributive property
(a+b)(3a-b)=3a^2-ab-3ab-b^2=3a^2-4ab-b^2
step 2
Multiply (3a^2-4ab-b^2) by (2a+7b)
6a^3+21a^2b-8a^2b-28ab^2-2ab^2-7b^3=6a^3+13a^2b-30ab^2-7b^3
answer is
<h2>6a^3+13a^2b-30ab^2-7b^3</h2>
Answer:
A) 2x³+11x²+8x-16
Step-by-step explanation:
When you multiply s(x) by t(x) you get something like this:
F(x) = x/2
g(x) = x - 3
f(8) = 8/2 = 4
g(f(x)) = 4 - 3 = 1
Answer g(f(8)) = 1
Answer:
m<GFA = 110
Step-by-step explanation:
1. ABCD - parallelogram Definition of a parallelogram
(AB ll CD) (AD ll BC)
2. m<B + m<C = 180 Consectuive angles in a
110 + m<C = 180 parallelogram are supplementary
m<C = 70
3. m<GCB = 1/2 m<C Definition of angles bisector
m<GCB = 70
4. m<B = m<D = 110 Opposite angles in a
parrallelogram are congruent
5. m<CDG = 1/2 m<D Defintion of an angle bisector
m<CDG = 55
6. m<GCB+m<CDG+m<CGD=180 Sum of anlges in a triangle (ΔCDG)
70 + 55 + m<CGD = 180
125 + m<CGD = 180
m<CGD = 55
7. m<CGD + m<DGF = 180 Linear pair, supplmentary angles
55 + m<DGF = 180
m<DGF = 125
8. m<C = m<A = 70 Opposite angles in a paralellogram
are congruent
9. m<ADG = 1/2m<D Definiton of an angle bisector
m<ADG = 55
10.m<ADG+m<DFG+m<GFA+m<A=360 Sum of angles in quadrilateral
55 + 125 + m<GFA + 70 = 360 DGFA
m<GFA + 250 = 360
m<GFA = 110
