Answer:
Now, 3x 2+x+5⩾0
This is because b² −4ac=1−4×5×3
=−59 (roots are imaginary)
3x²+x+5=(x−3)² =x²+9−6x and x−3⩾0
2x²+7x−4=0
2x²+8x−x−4=0
2x(x+4)−1(x+4)=0
(2x−1)(x+4)=0
x= 1/2 and−4 but x≥3
∴ No solution.
lol hehehe
Do not go to that link it is a virus and if you go to “goggle” and press the camera icon on the top right and take a picture of the question you should get the answer there
Let x = cost of the dishwasher.
The amounts the children are paying are:
Mark =20% = 1/5, so 1/5x
Sorin = 75
Raymond = 1/4x
Stephanie = 75
Kristin = 1/3x
Add them together:
x = 1/5x + 1/4x + 1/3x + 75 + 75
x - ( 1/5x + 1/4x + 1/3x ) = 150
Rewrite fractions with common denominator:
x - ( 12/60x + 15/60x + 20/60x ) = 150
60/60x - 47/60x = 150
13/60x = 150
x = 150/1 • 60/13
x = 9000/13
x = 692.31
The total cost of the dishwasher is $692.31
Now calculate the amount each child pays:
Mark = 1/5 x = 692.31/ 5 = $138.46
Sorin = $75.00
Raymond = 1/4 x = 692.31 / 4 = $173.08
Stephanie = $75.00
Kristin = 1/3 x = 692.31 / 3 = $230.77
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:
-2
Step-by-step explanation:
intercept at (0, -2)
