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Lilit [14]
3 years ago
14

three players scored a total of 70 points during a basketball game one player scored twice as many as one mate and 10 points few

er than the other teammate. how many points did each player score
Mathematics
2 answers:
Fed [463]3 years ago
7 0
40+20+10=70
40 is twice 20
10 is 10 fewer than 20.
Naily [24]3 years ago
3 0

Answer:

24, 12 and 34.

Step-by-step explanation:

Let points of all three players are x_1,\ x_2,\ x_3 respectively.

It is given that sum of their score is 70.

Therefore, x_1+x_2+x_3=70. ...1

Also, score of one is twice the other.

So, x_1=2x_2.   ...2

It is also given that its score is 10 points less than the other one.

Therefore, x_1=x_3-10  ...3

Solving all three equations we get,

x_1=24, \ x_2=12,\ x_3=34.

Hence,it is the required solution.

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Write the expression:<br><br> The quotient of the sum 3y and 5, and y squared.
lesya692 [45]

Answer:

(3y + 5) / (y^2)

Step-by-step explanation:

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6 0
2 years ago
Factor completely and find the roots of the following . X^2-6x+8=0
LiRa [457]

Answer:

x =2, x = 4

Step-by-step explanation:

x^2 - 6x + 8 = 0

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x - 2 = 0; x = 2

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5 0
2 years ago
Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
fenix001 [56]

Answer:

2, 7, 114

Step-by-step explanation:

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-2x² + 7x + 114

7 0
3 years ago
To change a recipe that serves 8 to make it serve 5, by what fraction would you multiply each ingredient?
NARA [144]
Let, that fraction = x
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8 0
3 years ago
Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)
AlekseyPX

Answer:

The approximate solution to the system is (-2.7, 2.1).

Step-by-step explanation:

To solve the system of equations \begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix} you must:

\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}

\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}

\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7

\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1

The approximate solutions to the system of equations are:

x=-2.7 ,\:y=2.1

5 0
3 years ago
Read 2 more answers
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