Answer:
Kenny has eaten <u>44 ice cream scoops</u>.
Step-by-step explanation:
Given:
Kenny uses 
of the chocolate syrup bottle for every scoop of ice cream that he eats.
Now, to find the scoops of ice cream Kenny has eaten, If he has used 
bottles of chocolate syrup.
So, we use unitary method to find the number of scoops of ice cream:
<em>If, </em><em> of the chocolate syrup bottle Kenny uses to eat = </em>
<em />
<em>Then, 1 of the chocolate syrup bottle Kenny uses to eat </em>
<em />
<em>Thus, </em><em> bottles of chocolate syrup bottle Kenny uses to eat</em> = 
Therefore, Kenny has eaten 44 ice cream scoops.
Yes, the above statement is true.
2×4x=8x plus 2×3y=6y.
The difference in the rate of change (RTC) would be 0.2.
The graph has a Slope of 0.4 and the table has a Slope of 0.2
Then you must subtract to find the difference
0.4 - 0.2 = 0.2
Hope this helped!
This is tricky. Fasten your seat belt. It's going to be a boompy ride.
If it's a 12-hour clock (doesn't show AM or PM), then it has to gain
12 hours in order to appear correct again.
How many times must it gain 3 minutes in order to add up to 12 hours ?
(12 hours) x (60 minutes/hour) / (3 minutes) = 240 times
It has to gain 3 minutes 240 times, in order for the hands to be in the correct positions again. Each of those times takes 1 hour. So the job will be complete in 240 hours = <em>10 days .</em>
Check:
In <u>10</u> days, there are <u>240</u> hours.
The clock gains <u>3</u> minutes every hour ==> <u>720</u> minutes in 240 hours.
In 720 minutes, there are 720/60 = <u>12 hours</u> yay !
_________________________________
If you are on a military base and your clocks have 24-hour faces,
then at the same rate of gaining, one of them would take 20 days
to appear to be correct again.
_________________________________
Note:
It doesn't have to be an analog clock. Cheap digital clocks can
gain or lose time too (if they run on a battery and don't reference
their rate to the 60 Hz power that they're plugged into).
Answer:
The set of solutions is ![\{\left[\begin{array}{c}a\\b\\x\\y\\z\end{array}\right] = \left[\begin{array}{c}-26+503y+543z\\-37+655y+724z\\-4+80y+90z\\y\\z\end{array}\right] : \text{y, z are real numbers}\}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5C%5Cb%5C%5Cx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-26%2B503y%2B543z%5C%5C-37%2B655y%2B724z%5C%5C-4%2B80y%2B90z%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3A%20%5Ctext%7By%2C%20z%20are%20real%20numbers%7D%5C%7D)
Step-by-step explanation:
The matrix associated to the problem is ![A=\left[\begin{array}{ccccc}1&-1&2&-8&1\\2&-1&-4&1&-2\\-4&1&4&-3&-1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-1%262%26-8%261%5C%5C2%26-1%26-4%261%26-2%5C%5C-4%261%264%26-3%26-1%5Cend%7Barray%7D%5Cright%5D)
and the vector of independent terms is (3,1,-1)^t. Then the matrix equation form of the system is Ax=b.
The vector equation form is ![a\left[\begin{array}{c}1\\2\\-4\end{array}\right]+b\left[\begin{array}{c}-1\\-1\\1\end{array}\right] + x\left[\begin{array}{c}2\\-4\\4\end{array}\right]+y\left[\begin{array}{c}-8\\1\\-3\end{array}\right] + z\left[\begin{array}{c}1\\-2\\-1\end{array}\right]=\left[\begin{array}{c}3\\1\\-1\end{array}\right]](https://tex.z-dn.net/?f=a%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%2Bb%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%5C%5C-1%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2B%20x%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-4%5C%5C4%5Cend%7Barray%7D%5Cright%5D%2By%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-8%5C%5C1%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20%2B%20z%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C-2%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%5C%5C1%5C%5C-1%5Cend%7Barray%7D%5Cright%5D)
.
Now we solve the system.
The aumented matrix of the system is ![\left[\begin{array}{cccccc}1&-1&2&-8&1&3\\2&-1&-4&1&-2&1\\-4&1&4&-3&-1&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccccc%7D1%26-1%262%26-8%261%263%5C%5C2%26-1%26-4%261%26-2%261%5C%5C-4%261%264%26-3%26-1%26-1%5Cend%7Barray%7D%5Cright%5D)
.
Applying rows operations we obtain a echelon form of the matrix, that is ![\left[\begin{array}{cccccc}1&-1&2&-8&1&3\\0&1&-8&-15&-4&-5\\0&0&1&-80&-9&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccccc%7D1%26-1%262%26-8%261%263%5C%5C0%261%26-8%26-15%26-4%26-5%5C%5C0%260%261%26-80%26-9%26-4%5Cend%7Barray%7D%5Cright%5D)
Now we solve for the unknown variables:
- x-80y-90z=-4 then x=-4+80y+90z
- b-8x-15y-4z=-5, b-8(-4+80y+90z)-15y-4z=-5 then b=-37+655y+724z.
- a-b+2x-8y+z=3, a-(-37+655y+724z)+2(-4+80y+90z)-8y+z=3, then a=-26+503y+543z
Since the system has two free variables then has infinite solutions.
The set of solutions is
