<h2>
x = 9</h2><h3 />
<u>We know:</u>
- we can add/subtract the same expression to both sides of the equation to get an equivalent equation;
- we can multiply/divide both sides of the equation by the same non-zero number to get the equivalent equation.
<u>Equivalent equations</u> are algebraic equations that have identical solutions.
We have the equation:
<h3>SOLUTION:</h3>
<em>multiply bot sides by 3</em>
<em />
<em>divide both sides by 7</em>
<em />
Answer: explanation
D SORRY IF IM WRONGGG AAAHHH
Answer:
I would say the Pennington line is steeper because the slope of the purple line is 2 and the other is 3/2 (don't know if they want slope or not) and it represents that it took 2.5 to three hours to travel 240 miles compared to the other one that takes four hours to travel the same distance. Hopefully that's right
Step-by-step explanation:
Answer:
No Solution.
Step-by-step explanation:
2x+3y=4
4x+6y=9
-------------
-2(2x+3y)=-2(4)
4x+6y=9
----------------------
-4x-6y=-8
4x+6y=9
-------------------
0=1
no solution
Answer:
t = 20.3 years
Step-by-step explanation:
I am assuming that this amount of money invested is compounding annually, so I am going to use the formula that goes along with that assumption:
where A(t) is the amount at the end of compounding, P is the initial investment, r is the interest rate in decimal form, and t is the time in years. We are solving for t. Right now it is the exponent, but we have to get it down from that position in order to solve for it. The only way we can do that is to eventually take the natural log of both sides. But let's write the equation first and then do some simplifying to make things a bit easier mathematically:
and
We will divide both sides by 5,500:
Taking the natural log of both sides gives us:
The power rule for logs (both common and natural) tells us that once we take the log or ln of a base, the exponent comes down out front:
ln(3.58181818) = t ln(1.065)
Now we can divide both sides by ln(1.065) and do the math on our calculators to find that
t = 20.2600 or, to the tenth of a year,
t = 20.3 years