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Slav-nsk [51]
3 years ago
13

In basketball, the number of points you score is given by 3x + 2y + 1z3x+2y+1z where xx is three-pointers, yy is two-pointers, a

nd zz is free throws (one-pointers). What is the total number of points scored by a player who makes 33 three-pointers, 55 two-pointers, and 66 free throws?
Mathematics
2 answers:
viva [34]3 years ago
7 0

Answer: 25 points

Step-by-step explanation:

Here, the expression that shows the total number of points scored by a player is,

3 x + 2 y + 1 z

Where, x = three-pointer,

y = two-pointer

z = one-pointer,

Here, x = 3, y= 5 and z = 6,

Hence, the total number of points = 3 x + 2 y + 1 z = 3 × 3 + 2 × 5 + 1 × 6 = 9 + 10 + 6 = 25

ExtremeBDS [4]3 years ago
6 0

Answer:

The total number of points scored by the player is 275.

Step-by-step explanation:

The number of points are represented by the expression is

T=3x+2y+1z

Where, x is three-pointers, y is two-pointers, and z is free throws.

It is given that a player makes 33 three-pointers, 55 two-pointers, and 66 free throws. The total number of points scored by that player is

T=3(33)+2(55)+1(66)

T=99+110+66

T=275

Therefore the total number of points scored by the player is 275.

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Read 2 more answers
What is the solution to the following system?
leva [86]

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]

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]

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]

\frac{R_2}{9}=R_2


\left[\begin{array}{cccc}1&-4&-1&-10\\0&1&\frac{4}{9}&\frac{35}{9}\\0&14&4&50\end{array}\right]


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]


-\frac{9}{20}R_3=R_3

\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]


-\frac{4}{9}R_3+R_2=R2 , -\frac{7}{9}R_3+R_1=R_1


\left[\begin{array}{cccc}1&0&0&4\\0&1&0&3\\0&0&1&2\end{array}\right]


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.

7 0
4 years ago
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