Subtracting the second equation by 18 on both sides, we have xy=-18. Next, we divide both sides by x to get y=-18/x Plugging that into the first equation, we have x+2(-18/x)=9. Multiplying both sides by x, we get x^2-36=9x. After that, we subtract both sides by 9x to get x^2-9x-36=0. Finding 2 numbers that add up to -9 but multiply to -36, we do a bit of guess and check to find the answers to be -12 and 3. Factoring it, we get
x^2-12x+3x-36=x(x-12)+3(x-12)=(x+3)(x-12). To find the x values, we have to find out when 0=(x+3)(x-12). This is simple as when you multiply 0 with anything, it is 0. Therefore, x=-3 and 12. Plugging those into x=-18/y, we get x=-18/y and by multiplying y to both sides, we get xy=-18 and then we can divide both sides by x to get -18/x=y. Plugging -3 in, we get -18/-3=6 and by plugging 12 in we get -18/12=-1.5. Therefore, our points are (-3,6) and (12, -1.5)
He traveled 2080 miles after 4 hours. He covered 3640 miles in 7 hours with a constant speed, so we can calculate the speed: 3640/7=520 miles/hour. After 4 hours, he traveled 520*4=2080 miles.
Answer:
Explanation:
Translate every verbal statement into an algebraic statement,
<u>1. Keith has $500 in a savings account at the beginning of the summer.</u>
<u>2. He wants to have at least $200 in the account by the end of summer. </u>
<u />
<u>3. He withdraws $25 a week for his cell phone bill.</u>
<u />
- Call w the number of weeks
<u>4. Write an inequality that represents Keith's situation.</u>
- Create your model: Final amount = Initial amount - withdrawals ≥ 500
With that inequality you can calculate how many week will pass before his account has less than the amount he wants to have in the account by the end of summer:
That represents that he can afford spending $ 25 a week during 12 weeks to have at least $ 200 in the account.
Answer:
all work is shown/pictured
Answer:
Step-by-step explanation:
Let the integer be 6 for even and 7 for odd (say)
For 6, we divide by 2, now get 3. Now we multiply by 3 and add 1 to get 10. Now since 10 is even divide by 5, now multiply by 3 and add 1 to get 16. Now divide by 2 again by 2 again by 2 again by 2 till we get rid of even numbers.
The result is 1, so multiply by 3 and add 1 we get 4 now divide 2 times by 2 to get 1, thus this result now again repeats after 2 times.
Say if we select off number 3, multiply by 3 and add 1 to get 10 now divide by 5, now repeat the same process as above for 5 until we get 1 and it gets repeated every third time.
Thus whether odd or even after some processes, we get 1 and the process again and again returns to 1.
