Answer:
20,21
Step-by-step explanation:
Let x be the first number.
Let y be the 2nd number.
Given y - x = 1 (since they are consecutive)
Rearrange it: y = x+1 (equation 1)
Also given:
(equation 2)
Now we can substitute y = x + 1 into equation 2.
Since we know x, the smaller number is 20, we can substitute to equation 1 to find y, the larger number.
y = 20+1
= 21.
Therefore the 2 numbers are 20 and 21.
Answer:
7/16
Approximately 0.44
Step by step explanation:
For the 1st student to arrive before 2nd student,
* the probability that he(1st student)to arrive during 3/4 hrs = 3/4
*Then he(1st student) will need to wait for 1/4 hrs
*The probability that he(1st student) arrive during the last 1/4 hrs = 1/4
*Then he(1st student) waits for average of 1/8 hrs (1/4 + 1/4)=1/8
*The total time he(1st student) wait = (3/4 ×1/4) +(1/4× 1/8) = 7/32
*For 1st and 2nd student to meet, the 2nd student need to arrive when 1st student is still arround, therefore the probability that 2nd student arrives when 1st student is arround = 7/32
*If 2nd student arrives before 1st student, the probability that they would meet is 7/32
* Adding both together 7/32 + 7/32 = 7/16
ANSWER = 0.4375
=0.44
The area of the circular sector is
From that, the equation for the central angle is
The perimeter of the circular sector is the sum of the arc length and twice the radius.
From that, the equation for the central angle is
The value of r can be found by equating these two expressions for θ.
By any of several means, you can find that r=0.2 or 1.25. The latter value corresponds to the requirement that the central angle be less than 90°. The radius is ...
1.25 miles.
The answer is A
because there is no other answer
Answer:
The answer is below
Step-by-step explanation:
Let s(t) represent the amount of salt in the tank at time t. Therefore:
ds / dt = (rate of salt flowing into the tank) - (rate of salt going out of the tank)
![\frac{ds}{dt}=[(0.08\ kg/L*6\ L/min)+(0.09\ kg/L*5\ L/min)]-(\frac{s\ kg}{1000\ L}*11\ L/min ) \\\\\frac{ds}{dt}=(0.48\ kg/min+0.45\ kg/min)-(\frac{11s}{1000}\ kg/min )\\\\\frac{ds}{dt}=0.93\ kg/min-\frac{11s}{1000}\ kg/min \\\\\frac{ds}{dt}=\frac{930-11s}{1000}\ kg/min \\\\\frac{ds}{930-11s}=\frac{1}{1000}dt\\\\Integrating:\\\\\int\limits { \frac{ds}{930-11s}} \, ds=\int\limits {\frac{1}{1000}} \, dt\\\\-\frac{1}{11}ln(930-11s)=\frac{t}{1000}+C\\\\multiply\ through\ by\ -11:](https://tex.z-dn.net/?f=%5Cfrac%7Bds%7D%7Bdt%7D%3D%5B%280.08%5C%20kg%2FL%2A6%5C%20L%2Fmin%29%2B%280.09%5C%20kg%2FL%2A5%5C%20L%2Fmin%29%5D-%28%5Cfrac%7Bs%5C%20kg%7D%7B1000%5C%20L%7D%2A11%5C%20L%2Fmin%20%29%20%5C%5C%5C%5C%5Cfrac%7Bds%7D%7Bdt%7D%3D%280.48%5C%20kg%2Fmin%2B0.45%5C%20kg%2Fmin%29-%28%5Cfrac%7B11s%7D%7B1000%7D%5C%20kg%2Fmin%20%29%5C%5C%5C%5C%5Cfrac%7Bds%7D%7Bdt%7D%3D0.93%5C%20kg%2Fmin-%5Cfrac%7B11s%7D%7B1000%7D%5C%20kg%2Fmin%20%5C%5C%5C%5C%5Cfrac%7Bds%7D%7Bdt%7D%3D%5Cfrac%7B930-11s%7D%7B1000%7D%5C%20kg%2Fmin%20%5C%5C%5C%5C%5Cfrac%7Bds%7D%7B930-11s%7D%3D%5Cfrac%7B1%7D%7B1000%7Ddt%5C%5C%5C%5CIntegrating%3A%5C%5C%5C%5C%5Cint%5Climits%20%7B%20%5Cfrac%7Bds%7D%7B930-11s%7D%7D%20%5C%2C%20ds%3D%5Cint%5Climits%20%7B%5Cfrac%7B1%7D%7B1000%7D%7D%20%5C%2C%20%20dt%5C%5C%5C%5C-%5Cfrac%7B1%7D%7B11%7Dln%28930-11s%29%3D%5Cfrac%7Bt%7D%7B1000%7D%2BC%5C%5C%5C%5Cmultiply%5C%20through%5C%20by%5C%20-11%3A)
