4 2 as a fraction because your doing the unshaded and the shaded part of the ratio so that would be in a fraction to 4 and 2 as a fraction
Answer:
The largest possible volume V is ;
V = l^2 × h
V = 20^2 × 10 = 4000cm^3
Step-by-step explanation:
Given
Volume of a box = length × breadth × height= l×b×h
In this case the box have a square base. i.e l=b
Volume V = l^2 × h
The surface area of a square box
S = 2(lb+lh+bh)
S = 2(l^2 + lh + lh) since l=b
S = 2(l^2 + 2lh)
Given that the box is open top.
S = l^2 + 4lh
And Surface Area of the box is 1200cm^2
1200 = l^2 + 4lh ....1
Making h the subject of formula
h = (1200 - l^2)/4l .....2
Volume is given as
V = l^2 × h
V = l^2 ×(1200 - l^2)/4l
V = (1200l - l^3)/4
the maximum point is at dV/dl = 0
dV/dl = (1200 - 3l^2)/4
dV/dl = (1200 - 3l^2)/4 = 0
3l^2= 1200
l^2 = 1200/3 = 400
l = √400
I = 20cm
Since,
h = (1200 - l^2)/4l
h = (1200 - 20^2)/4×20
h = (800)/80
h = 10cm
The largest possible volume V is ;
V = l^2 × h
V = 20^2 × 10 = 4000cm^3
Answer:
Sister = 15 years old
Brother = 12 years old
Step-by-step explanation:
Right now, if the brother is B years old and the sister is S years old, B = (4/5)S because the ratio from brother to sister is 4:5, so if one unit is X, B = 4X and S = 5X
Three years ago, the brother was B-3 years old and the sister was S-3 years old. Keeping one unit as X, we have
B = 3X
S = 4X
B-3 = (3/4)(S-3)
Therefore, we have a system of equations
B = (4/5)S
B-3 = (3/4)(S-3)
Substitute (4/5)S for B into the second equation to only have one variable
(4/5)S - 3 = (3/4)(S-3)
(4/5)S - 3 = (3/4)S - 9/4
add 3 to both sides and subtract (3/4)S from both sides to isolate the variable and its coefficient
(1/20)S = 3/4
multiply both sides by 20 to isolate the S
S = 15
B = (4/5)S = 12
Answer:
$6.21
Step-by-step explanation:
1 apple = $0.60
3 apples = $1.80 (Multiply 0.60 by 3)
2 bananas = $1.00 (Multiply 0.50 by 2)
1 quart of yoghurt = $4.50
Add the prices up first:
1.80 + 1.00 + 4.50 = $7.30
7.30 off 15% = $6.21
The answer is negative, because the first number is negative.
