Answer:
(4,3,2)
Step-by-step explanation:
We can solve this via matrices, so the equations given can be written in matrix form as:
![\left[\begin{array}{cccc}3&2&1&20\\1&-4&-1&-10\\2&1&2&15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%262%261%2620%5C%5C1%26-4%26-1%26-10%5C%5C2%261%262%2615%5Cend%7Barray%7D%5Cright%5D)
Now I will shift rows to make my pivot point (top left) a 1 and so:
![\left[\begin{array}{cccc}1&-4&-1&-10\\2&1&2&15\\3&2&1&20\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C2%261%262%2615%5C%5C3%262%261%2620%5Cend%7Barray%7D%5Cright%5D)
Next I will come up with algorithms that can cancel out numbers where R1 means row 1, R2 means row 2 and R3 means row three therefore,
-2R1+R2=R2 , -3R1+R3=R3
![\left[\begin{array}{cccc}1&-4&-1&-10\\0&9&4&35\\0&14&4&50\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C0%269%264%2635%5C%5C0%2614%264%2650%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-4&-1&-10\\0&1&\frac{4}{9}&\frac{35}{9}\\0&14&4&50\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%2614%264%2650%5Cend%7Barray%7D%5Cright%5D)
4R2+R1=R1 , -14R2+R3=R3
![\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&-\frac{20}{9}&-\frac{40}{9}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%5Cfrac%7B7%7D%7B9%7D%26%5Cfrac%7B50%7D%7B9%7D%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%260%26-%5Cfrac%7B20%7D%7B9%7D%26-%5Cfrac%7B40%7D%7B9%7D%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%5Cfrac%7B7%7D%7B9%7D%26%5Cfrac%7B50%7D%7B9%7D%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%260%261%262%5Cend%7Barray%7D%5Cright%5D)
, 
![\left[\begin{array}{cccc}1&0&0&4\\0&1&0&3\\0&0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%264%5C%5C0%261%260%263%5C%5C0%260%261%262%5Cend%7Barray%7D%5Cright%5D)
Therefore the solution to the system of equations are (x,y,z) = (4,3,2)
Note: If answer choices are given, plug them in and see if you get what is "equal to". Meaning plug in 4 for x, 3 for y and 2 for z in the first equation and you should get 20, second equation -10 and third 15.
A) The average market price is (12 + 8 + 10 + 13 + 14 + 8) / 6 = $10.83
b)
The total amount spent was: 12*8 + 8*12 + 10*10 + 13*7 + 14*7 + 8*12 =
$577. The total number of shares was: 8 + 12 + 10 + 7 + 7 + 12 = 56.
Therefore the average price per share was $577 / 56 shares =
$10.30/share.
c) Since the cost per share in June was $12, and her average cost was $10.30, dollar-cost averaging worked in her favor since her average cost was lower.
Answer:
1. m∠B=110°
2. 560 cm3
3. Numerical data
4. 2000 cm3
5. 50%
Step-by-step explanation:
1. The explanation of part 1 is given in the attachment.
2. Given dimensions : 10 cm, 8 cm, and 7 cm.
Let Length of cuboid =10 cm
breadth/width of cuboid =8 cm
height of cuboid = 7cm
Volume of cuboid = length *width* height
=( 10 *8*7) cm3
=(560) cm3
3. Age, Birth date and weight are the types/examples of "<u>Numerical Data"</u> because these all are describe in terms of numeric values.
4. 1 liter = 1000 cm3 or 1 cm3 = 0.001 liter
1.5 liters =(1.5*1000) cm3 = (15*100) =1500 cm3
1 dm3 =1000 cm3
0.35 dm3 = (0.35*1000) cm3 = (35*10) cm3 =350 cm3
Given expression: 1.5 litre + 0.35 dm3 + 150 cm3 = <u> </u> cm3
1500 cm3 + 350 cm3 +150 cm3 = <u>2000</u> cm3
5. If A=(1/2)B, then B : A = <u>50</u> %
Ratio: B : A
B : (1/2) B
1: (1/2)
50 % (The value of A is half of the value of B)
9/10....................................
Answer:
24c
Step-by-step explanation:
I see 4 times 6c, which is,
24c
Hope this helps!!
