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Dmitry_Shevchenko [17]
3 years ago
15

Si un jugador de baloncesto encesta 8 lanzamientos 12 intentos, ¿cuántos tienen que hacer para lograr 20 canastas mantener el mi

smo nivel de acierto?
Mathematics
1 answer:
n200080 [17]3 years ago
6 0
For this case we have the following rule of three:
 12 ----> 100%
 8 ------> x
 Clearing x we have:
 x = (8/12) * (100)
 x = 66.66666667% (percentage of correct answers)
 For 20 baskets we have:
 20 -----> 66.66666667%
 x -------> 100%
 Clearing x we have:
 x = (100 / 66.66666667) * (20)
 x = 30 attempts
 Answer:
 
To make 20 canatas, 30 attempts must be made.
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URGENT WILL GIVE BRAINLIEST
elena-s [515]

Answer:

(5sqrt(2), 45 deg)

(-5sqrt(2), 225 deg)

Step-by-step explanation:

(x,y)=(5,5)

So theta=arctan(5/5)=arctan(1)=45 degrees

Now r! r=sqrt(x^2+y^2)=sqrt(5^2+5^2)=sqrt(50)=sqrt(25)sqrt(2)=5sqrt(2)

So one polar point is (5sqrt(2) , 45 degrees)

Now if we do 180+45=225 degrees... this puts us in the 3rd quadrant... to get back to quadrant 1 we just take the opposite of our r

so another point is (-5sqrt(2) , 225 degrees)

7 0
3 years ago
Which proportion models this problem?
Ilya [14]
Thank you for posting your question here at brainly. Feel free to ask more questions.  

The best and most correct answer among the choices provided by the question is <span>C. x over 520 = 5 over 13</span>.<span>       
    </span><span>
Hope my answer would be a great help for you.</span>
7 0
3 years ago
Rectangle 1 has length x and width y. Rectangle 2 is made by multiplying each dimension of Rectangle 1 by a factor of K, where k
STALIN [3.7K]

\bold{\huge{\underline{\pink{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

  • <u>Rectangle </u><u>1</u><u> </u><u> </u><u>has </u><u>length </u><u>x</u><u> </u><u>and </u><u>width </u><u>y</u>
  • <u>Rectangle</u><u> </u><u>2</u><u> </u><u>is </u><u>made </u><u>by </u><u>multiplying </u><u>each </u><u>dimensions </u><u>of </u><u>rectangle </u><u>1</u><u> </u><u>by </u><u>a </u><u>factor </u><u>of </u><u>k </u>
  • <u>Where</u><u>, </u><u>k </u><u>></u><u> </u><u>0</u><u> </u><u> </u>

<h3><u>Answer </u><u>1</u><u> </u><u>:</u><u>-</u></h3>

Yes, The rectangle 1 and rectangle 2 are similar .

<h3><u>According </u><u>to </u><u>the </u><u>similarity </u><u>theorem </u><u>:</u><u>-</u></h3>

  • If the ratio of length and breath of both the triangles are same then the given triangles are similar.

<u>Let's </u><u>Understand </u><u>the </u><u>above </u><u>theorem </u><u>:</u><u>-</u>

The dimensions of rectangle 1 are x and y

<u>Now</u><u>, </u>

  • Rectangle 2 is made by multiplying each dimensions of rectangle 1 by a factor of k .

Let assume the value of K be 5

<u>Therefore</u><u>, </u>

The dimensions of rectangle 2 are

\sf{ 5x \:and \:5y }

<u>Now</u><u>, </u><u> </u><u>The </u><u>ratios </u><u>of </u><u>dimensions </u><u>of </u><u>both </u><u>the </u><u>rectangle </u><u>:</u><u>-</u>

  • \bold{Rectangle 1 =  Rectangle 2}

\bold{\dfrac{ x }{y}}{\bold{ = }}{\bold{\dfrac{5x}{5y}}}

\bold{\blue{\dfrac{ x }{y}}}{\bold{\blue{ = }}}{\bold{\blue{\dfrac{x}{y}}}}

<u>From </u><u>above</u><u>, </u>

We can conclude that the ratios of both the rectangles are same

Hence , Both the rectangles are similar

<h3><u>Answer </u><u>2</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>Here</u><u>, </u>

We have to proof that, the

  • Perimeter of rectangle 2 = k(perimeter of rectangle 1 )

In the previous questions, we have assume the value of k = 5

<h3><u>Therefore</u><u>, </u></h3>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

Perimeter of rectangle 1

\sf{ = 2( length + Breath) }

\bold{\pink{= 2( x + y ) }}

Thus, The perimeter of rectangle 1

Perimeter of rectangle 2

\sf{ = 2( length + Breath) }

\sf{ = 2(5x + 5y) }

\sf{ =  2 × 5( x + y) }

\bold{\pink{= 10(x + y) }}

Thus, The perimeter of rectangle 2

<u>According </u><u>to </u><u>the </u><u>given </u><u>condition </u><u>:</u><u>-</u>

  • Perimeter of rectangle 2 = k( perimeter of rectangle 1 )

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 2(x + y) = 10(x + y)}

\bold{\pink{2x + 2y = 5(2x + 2y) }}

<u>From </u><u>Above</u><u>, </u>

We can conclude that the, Perimeter of rectangle 2 is 5 times of perimeter of rectangle 1 and we assume the value of k = 5.

Hence, The perimeter of rectangle 2 is k times of rectangle 1

<h3><u>Answer 3 :</u></h3>

<u>Here</u><u>, </u>

We have to proof that ,

  • <u>The </u><u>area </u><u>of </u><u>rectangle </u><u>2</u><u> </u><u>is </u><u>k²</u><u> </u><u>times </u><u>of </u><u>the </u><u>area </u><u>of </u><u>rectangle </u><u>1</u><u>.</u>

<u>That </u><u>is</u><u>, </u>

  • Area of rectangle 1 = k²( Area of rectangle)

<h3><u>Therefore</u><u>, </u></h3>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

<u>Area </u><u>of </u><u>rectangle </u><u>1</u>

\sf{ = Length × Breath }

\sf{ = x × y }

\bold{\red{= xy }}

<u>Area </u><u>of </u><u>rectangle </u><u>2</u>

\sf{ = Length × Breath }

\sf{ = 5x × 5y }

\bold{\red{ = 25xy }}

<u>According </u><u>to </u><u>the </u><u>given </u><u>condition </u><u>:</u><u>-</u>

  • Area of rectangle 1 = k²( Area of rectangle)

\sf{ xy = 25xy }

\bold{\red{xy = (5)²xy }}

<u>From </u><u>Above</u><u>, </u>

We can conclude that the, Area of rectangle 2 is (5)² times of area of rectangle 1 and we have assumed the value of k = 5

Hence, The Area of rectangle 2 is k times of rectangle 1 .

7 0
2 years ago
What's the formula of<br><br><br>(a+b)³ = ?<br><br>​
zzz [600]

Answer:

(a+b)³=a³+3a²b+3ab²+b³

6 0
3 years ago
Read 2 more answers
Given the expression-1/2(4x-6) +2 (x2-5x+5) complete the instructions below
Art [367]

Answer:

  A) 2x^2 -12x +13

  B) terms {2x^2, -12x, 13}; coefficients {2, -12, 13}; constants {13}

Step-by-step explanation:

A) Using the distributive property, we get ...

  -2x +3 +2x^2 -10x +10

  = 2x^2 -12x +13

__

B) Terms: {2x^2, -12x, 13}

  Factors: There are no overall factors. Each term has factors ...

     2x^2 has factors {2, x, x}

     -12x has factors {-1, 2, 2, 3, x}

     13 has factors {1, 13} (it is prime)

  Coefficients: {2, -12, 13} . . . see note below

  Constants: {13}

_____

<em>Additional comment</em>

We have listed the prime factors of each term, any subset can be multiplied to be a factor of the term. For example, 2·3 = 6 is a factor of -12x.

Please note that some sources do not consider a constant (13) to be a coefficient. You will need to refer to your curriculum materials to see what your author says about that.

4 0
3 years ago
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