Answer:
B) 4x^2y^2
Step-by-step explanation:
8x^3y^2+20x^2y^4
4(2x^3y^2+5x^2y^4)
4x^2(2xy^2+5y^4)
4x^2y^2(2x+5y^2)
So the answer is B
point slope form
y-y1 = m(x-x1)
y-9 = 2(x-1)
change to slope intercept form
y-9 = 2(x-1)
distribute
y-9 = 2x-2
add 9 to each side
y = 2x-2+9
y = 2x+7
T is the cost of the table. B is the cost of a bench.
b+t=652
b+98=t
t-98+t=652
2t=554
t=277
b=375
Hope this helps!
Answer:
d = k·sin(2θ)·sin(α)/(sin(θ)·sin(β))
Step-by-step explanation:
The Law of Sines tells us that sides of a triangle are proportional to the sine of the opposite angle. This can be used along with a trig identity to demonstrate the required relation.
__
<h3>top triangle</h3>
The law of sines applied to the top triangle is ...
BC/sin(A) = AC/sin(θ)
Triangle ABC is isosceles, so the base angles at B and C are congruent. Then the angle at vertex A is ...
∠A = 180° -θ -θ = 180° -2θ
A trig identity tells us the sine of an angle is equal to the sine of its supplement. That means the sine of angle A is ...
sin(A) = sin(180° -2θ) = sin(2θ)
and our above Law of Sines equation tells us ...
BC = sin(A)/sin(θ)·AC = k·sin(2θ)/sin(θ)
__
<h3>bottom triangle</h3>
The law of sines applied to the bottom triangle is ...
DC/sin(B) = BC/sin(D)
d/sin(α) = BC/sin(β)
Multiplying by sin(α) we have ...
d = BC·sin(α)/sin(β)
__
Using our expression for BC gives the desired relation:
d = k·sin(2θ)·sin(α)/(sin(θ)·sin(β))