2l+2w=100
2(w+20)+2w=100
2w+40+2w=100
4w+40=100
4w=60
w=15
l=w+20
l=15+20
Greatest length is 35ft.
Step-by-step explanation:
Any coordinates in the shaded area are solutions. Any coordinates on the line are not a solution due to it being dashed. Anything in the non-shaded area is also not a solution. You can always check if you are right by plugging in the numbers.
Example: (1, -2)
-2 < 0.5(1) + 2
-2 < 0.5 + 2
-2 < 2.5
True, -2 is less than 2.5.
Another Example: (-2, 1)
1 < 0.5(-2) + 2
1 < -1 + 2
1 < 1
False, 1 is not less than itself.
I use the sin rule to find the area
A=(1/2)a*b*sin(∡ab)
1) A=(1/2)*(AB)*(BC)*sin(∡B)
sin(∡B)=[2*A]/[(AB)*(BC)]
we know that
A=5√3
BC=4
AB=5
then
sin(∡B)=[2*5√3]/[(5)*(4)]=10√3/20=√3/2
(∡B)=arc sin (√3/2)= 60°
now i use the the Law of Cosines
c2 = a2 + b2 − 2ab cos(C)
AC²=AB²+BC²-2AB*BC*cos (∡B)
AC²=5²+4²-2*(5)*(4)*cos (60)----------- > 25+16-40*(1/2)=21
AC=√21= 4.58 cms
the answer part 1) is 4.58 cms
2) we know that
a/sinA=b/sin B=c/sinC
and
∡K=α
∡M=β
ME=b
then
b/sin(α)=KE/sin(β)=KM/sin(180-(α+β))
KE=b*sin(β)/sin(α)
A=(1/2)*(ME)*(KE)*sin(180-(α+β))
sin(180-(α+β))=sin(α+β)
A=(1/2)*(b)*(b*sin(β)/sin(α))*sin(α+β)=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
KE/sin(β)=KM/sin(180-(α+β))
KM=(KE/sin(β))*sin(180-(α+β))--------- > KM=(KE/sin(β))*sin(α+β)
the answers part 2) areside KE=b*sin(β)/sin(α)side KM=(KE/sin(β))*sin(α+β)Area A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
Answer:
I cant see the equation but it should go something along the lines of t = 30 + 4r
Step-by-step explanation:
t = total
r = number of rides
Example: If he went on 4 rides, hen u would imput 4 into r like this
t = 30 + 4 (4)
t = 30 + 16
t = 46
We are given to lines XY and VW. Now we need to determine the expression that correctly states that these lines are congruent. One possibility to prove that they're congruent is if they are two separate lines and:
XA is congruent to VB,
AY is congruent to BW
XA + AY = XY
VB + BW = VW
Then we can conclude that if the statements above are true, XY and VW must be congruent to each other.
Another possibility is that they are two sides of an isosceles rectangle XYVW and are opposite sides of the rectangle. <span />
