

- <u>We </u><u>have </u><u>given </u><u>two </u><u>linear </u><u>equations </u><u>that</u><u> </u><u>is </u><u>2x </u><u>-</u><u> </u><u>3y </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>and </u><u>x</u><u> </u><u>+</u><u> </u><u>3y </u><u>=</u><u> </u><u>1</u><u>2</u><u> </u><u>.</u>

- <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u><u>and </u><u>y </u><u>by </u><u>elimination </u><u>method</u><u>. </u>



<u>Multiply </u><u>eq(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>by </u><u>2</u><u> </u><u>:</u><u>-</u>


<u>Subtract </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>from </u><u>eq(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>





<u>Now</u><u>, </u><u> </u><u>Subsitute</u><u> </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>





Hence, The value of x and y is 2 and 10/3
He would maybe have 7.5 but I insist you the answer could be 8 too
Answer:
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Step-by-step explanation:
Given the data in the question;
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?
dA/dt = rate in - rate out
first we determine the rate in and rate out;
rate in = 3pound/gallon × 5gallons/min = 15 pound/min
rate out = A pounds/1000gallons × 5gallons/min = 5Ag/1000pounds/min
= 0.005A pounds/min
so we substitute
dA/dt = rate in - rate out
dA/dt = 15 - 0.005A
Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Answer:
3
Step-by-step explanation: