m□ebd=4 x-8 and m□ebc=5 x+20
This is solvable only if e b is the initial side and b d and b c lies on opposite side of each other and lies on a line i.e c,b,d are Collinear.
∠ebd and ∠ebc will form a linear pair.The meaning of linear pair is that angles forming on one side of a straight line through a common vertex which are adjacent is 180°.
i.e
∠ ebd + ∠ebc = 180°
4 x- 8 + 5x + 20= 180°
adding like terms
⇒ 9 x +12 =180°
⇒ 9 x = 180° - 12
⇒ 9 x = 168°
⇒ x =( 168/9)°=(56/3)°
now m□ebc =5 x +20
= 5 × 56/3 + 20
= 280/3 + 20
=340/3
m□ebc=( 340/3)°
So, solution set is x =(56/3)° and m□ebc =(340/3)°
9514 1404 393
Answer:
Step-by-step explanation:
The ratio of smallest sides in the two triangles is ...
smaller / larger = 4/10
To find A and B, we write proportions involving corresponding side lengths.
__
Side A in the smaller triangle is ...
A/15 = 4/10
A = (4/10)(15) = 6
__
The ratio of larger to smaller is the inverse of the above ratio, so ...
B/9 = 10/4
B = (10/4)(9) = 22.5
Answer:
Uche's pumping rate is <u>300 mL/s</u>.
Step-by-step explanation:
Given:
Uche pumps gasoline at a rate of 18 L/min.
Now, to find Uche's pumping rate in mL/s.
The rate at which Uche pumps gasoline = 18 L/min.
So, to get the pumping rate in mL/s we use convert L/min to mL/s by using conversion factor:
<u>1 L/min = 16.6667 mL/s.</u>
18 L/min = 16.6667 × 18 mL/s.
18L/min = 300 mL/s.
Therefore, Uche's pumping rate is 300 mL/s.
Answer:
Step-by-step explanation:
a) For a prime numbers we have array with 2 rectangulars R1: a=1 and b=prime number; R2: a=prime number and b=1. Both has the same are, that prime number.
b) For a composite number which are not square number we have rectanular array with even numbers of ractangulars. For example, number 6.
R1: a=1,b=6; R2: a=2,b=3; R3: a=3, b=2; R4: a=6,b=1. Each rectangular has the same area, 6.
c) The square number we alway have te odd number of rectanulars, because of the square a=x,b=x can not be simetric. For example 16.
R1: a=1,b=16; R2: a=2 , b=8; R3: a=4,b=4; R4: a=8, b=2; R5:a=16,b=1.Each rectangular has the same area, 16.
Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c
when t = 0, Q = 200 L × 1 g/L = 200 g
We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2
㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min
