Answer:
(A) ![A=\left[\begin{array}{ccc}10&20&40\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%2620%2640%5Cend%7Barray%7D%5Cright%5D)
(B) ![B=\left[\begin{array}{ccc}11&22&44\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%2622%2644%5Cend%7Barray%7D%5Cright%5D)
(C) ![A+B=\left[\begin{array}{ccc}21&42&84\end{array}\right]](https://tex.z-dn.net/?f=A%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%2642%2684%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The manager ordered 10 lb of tomatoes, 20 lb of zucchini, and 40 lb of onions from a local farmer one week.
(A)
Matrix <em>A</em> represents the amount of each item ordered. It is 1 × 3 matrix.
Then matrix <em>A</em> is:
![A=\left[\begin{array}{ccc}10&20&40\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%2620%2640%5Cend%7Barray%7D%5Cright%5D)
(B)
Next week the manager increases the order of all the products by 10%.
Then the amount of new orders are:
Tomatoes ![=10\times [1+\frac{10}{100}]=10\times1.10=11](https://tex.z-dn.net/?f=%3D10%5Ctimes%20%5B1%2B%5Cfrac%7B10%7D%7B100%7D%5D%3D10%5Ctimes1.10%3D11)
Zucchini ![=20\times [1+\frac{10}{100}]=20\times1.10=22](https://tex.z-dn.net/?f=%3D20%5Ctimes%20%5B1%2B%5Cfrac%7B10%7D%7B100%7D%5D%3D20%5Ctimes1.10%3D22)
Onions ![=40\times [1+\frac{10}{100}]=40\times1.10=44](https://tex.z-dn.net/?f=%3D40%5Ctimes%20%5B1%2B%5Cfrac%7B10%7D%7B100%7D%5D%3D40%5Ctimes1.10%3D44)
Th matrix <em>B</em> represents the amount of each order for the next week. Then matrix <em>B</em> is:
![B=\left[\begin{array}{ccc}11&22&44\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%2622%2644%5Cend%7Barray%7D%5Cright%5D)
(C)
Add the two matrix <em>A</em> and <em>B</em> as follows:
![A+B=\left[\begin{array}{ccc}10&20&40\end{array}\right]+\left[\begin{array}{ccc}11&22&44\end{array}\right]\\=\left[\begin{array}{ccc}(10+11)&(20+22)&(40+44)\end{array}\right]\\=\left[\begin{array}{ccc}21&42&84\end{array}\right]](https://tex.z-dn.net/?f=A%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%2620%2640%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%2622%2644%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2810%2B11%29%26%2820%2B22%29%26%2840%2B44%29%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%2642%2684%5Cend%7Barray%7D%5Cright%5D)
The entries of the matrix (<em>A</em> + <em>B</em>) represent the amount of tomatoes, zucchini and onions ordered for two weeks.
5 times 20(which is your number of days) equals 100. Louis already had 25 so add both numbers together to get 125
Answer:
the answer is the second choice
Niklas takes a dose of 25 micrograms of a certain supplement each day. The supplement has a half life of 4 hours, meaning that 1/64 of the supplement remains in the body after each day. How much of the supplement is in Niklas's body immediately after the 12th dose? Round your final answer to the nearest hundredth.
Answer:
The amount of the supplement in Niklas body immediately after the 12th dose is 430 micrograms to the nearest hundreth
Step-by-step explanation:
Half life is the time required for an element to decay into half of its initial size.
Given that :
The supplement has a half life of 4 hours, this implies that it decay to half of its size every 4 hours.
∴ there are 6 stages of division in a day.
i.e

The amount of the supplement in Niklas body after the first dose (first day) can be calculated as:
=
× 25
= 0.390625 micrograms
It is said that he used the supplement daily for 12 days (12th dose),
As such ; we can estimate the amount of the supplement that is in his body immediately after the 12th dose; which is calculated as:
amount in his body per day × number of period for complete decay
= 0.390625 × 11
= 4.296875
≅4.30 micrograms
= 430 micrograms to the nearest hundreth
The amount of the supplement in Niklas body immediately after the 12th dose is 430 micrograms to the nearest hundreth
Answer/Step-by-step explanation:
4(x - 2)= 100 and 2(x - 2) = 50.
We can determine that in both equation x equals 27. But, why is the second one half as much as the first problem? This is because the second problem is being multiplied with a number half as large as the first. Since 27 - 2 = 25 and 25 x 4 = 100 and 25 x 2 = 50 these problems hold up to their function.
If x = 27:
<u>4(x - 2) = 100</u>
27 - 2 = 25
25 x 4 = 100
<u>2(x - 2) = 50</u>
27 - 2 = 25
25 x 2 = 50