Answer: 3x-y=10 , 6x-2y=5. 3x−y=10 3 x - y = 10 , 6x−2y=5 6 x - 2 y = 5. Solve for y y in the first equation. Tap for more steps... Subtract 3x 3 x from
Graph y=6x-10 |
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:
p-e< p < p+e
(0.061 - 0.025) < 0.061 < (0.061 + 0.025)
0.036 < 0.061 < 0.086
Step-by-step explanation:
Given;
Confidence interval CI = (a,b) = (0.036, 0.086)
Lower bound a = 0.036
Upper bound b = 0.086
To express in the form;
p-e< p < p+e
Where;
p = mean Proportion
and
e = margin of error
The mean p =( lower bound + higher bound)/2
p = (a+b)/2
Substituting the values;
p = (0.036+0.086)/2
Mean Proportion p = 0.061
The margin of error e = (b-a)/2
Substituting the given values;
e = (0.086-0.036)/2
e = 0.025
Re-writing in the stated form, with p = 0.061 and e = 0.025
p-e< p < p+e
(0.061 - 0.025) < 0.061 < (0.061 + 0.025)
0.036 < 0.061 < 0.086
Sinα=h/L where h=height, L=string length...
h=Lsinα so
h(25°)=50sin25≈21.1ft
h(45°)=50sin45≈35.4ft
Answer:
option c is the correct answer
Step-by-step explanation:
∠YXZ+117°=180°
∠YXZ=180°-117°
∠YXZ=63°
63°+36°+∠YZX=180°(sum of angles is 180)
99°+∠YZX=180
∠YZX=180-99
∠YZX=81
