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mr_godi [17]
2 years ago
10

Lindsey Wilson was the top scorer in a women's professional basketball league for the 2006 regular season, with a total of 814 p

oints. The number of two-point field
goals that Lindsey made was 49 less than double the number of three-point field goals she made. The number of free throws (each worth one point) she made was

38 less than the number of two-point field goals she made. Find how many free throws, two-point field goals, and three-point field goals Lindsey Wilson made during

the 2006 regular season.

Lindsey Wilson made free throws.
Mathematics
1 answer:
baherus [9]2 years ago
3 0

Answer:

The number of free throws, two-point throw, and three-point throw are 135, 173, and 111 respectively.

Step-by-step explanation:

Let x,y, and z are the numbers of two-point field goals, numbers of three-point field goals, and the number of free-throws (one-point goal) respectively.

The total points= 814

\Rightarrow 2x+3y+z=814\;\cdots(i)

As the number of two-point  goals was 49 less than double the number of three-point field goals she made.

\Rightarrow x=2y-49 \;\cdots(ii)

Again, the number of free goals was 38 less than the number of two-point field goals she made.

\Rightarrow z=x-38\;\cdots(iii)

From equations (i) and (iii)

2x+3y+x-38=814

\Rightarrow 3x+3y=852

\Rightarrow x+y=284

\Rightarrow 2y-49+y=284 [using equation (ii)]

\Rightarrow 3y=333

\Rightarrow y=111

From equation (ii),

x= 2\times111-38=173

From equation (iii),

z=173-38=135.

Hence,

The number of free throws, z= 135

The number of two-point throw, x=173

The number of three-point throw, y=111.

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5 %

Step-by-step explanation:

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2 years ago
The Nutty Professor sells cashews for $7.10 per pound and Brazil nuts for $4.60 per pound. How much of each type should be used
aivan3 [116]

Answer:

Pounds for cashew- 20.96

Pounds for Brazilian nut- 11.04

Step-by-step explanation:

Let c be the pounds of cashew

Let b ne the pound of Brazilian but

32×6.24

= 199.68

7.1c + 4.60b= 199.68.......equation 1

c + b= 32..........equation 2

From equation 2

c= 32-b

Sub 32-b for c in equation 1

7.1(32-b) + 4.60b= 199.68

227.3-7.1b+4.60b, = 199.68

227.3-2.5b= 199.68

-2.5b= 199.69-227.3

- 2.5b= -27.61

b= 27.61/2.5

b= 11.04

Sub 11.04 for b in equation 2

c+b= 32

c + 11.04= 32

c= 32-11.04

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4 0
2 years ago
Solve for x in the equation x^2- 10x+ 25=35
DIA [1.3K]

Answer:

<h3>x = 5  +  \sqrt{35}  \:  \:  \:or \:  \:  \: x = 5 -  \sqrt{35}  \\</h3>

Step-by-step explanation:

x² - 10x + 25 = 35

Move 35 to the left side of the equation

That's

x² - 10x + 25 - 35 = 0

x ² - 10x - 10 = 0

Using the quadratic formula solve the equation

That's

<h3>x =  \frac{ - b\pm \sqrt{ {b}^{2} - 4ac } }{2a}</h3>

From the question

a = 1 , b = - 10 , c = - 10

Substitute the values into the above formula and solve

That's

x =  \frac{ -  - 10\pm \sqrt{( { - 10})^{2} - 4(1)( - 10) } }{2(1)}  \\  =  \frac{10\pm \sqrt{100 + 40} }{2}  \\  =  \frac{10\pm \sqrt{140} }{2}  \\  =  \frac{10\pm2 \sqrt{35} }{2}  \\  =  \frac{10}{2} \pm \frac{2 \sqrt{35} }{2}  \\  = 5\pm \sqrt{35}

We have the final answer as

<h3>x = 5  +  \sqrt{35}  \:  \:  \:or \:  \:  \: x = 5 -  \sqrt{35}  \\</h3>

Hope this helps you

7 0
2 years ago
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