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Marrrta [24]
3 years ago
15

Find the diameter of a circle with the given dimensions r=13 cm

Mathematics
2 answers:
maw [93]3 years ago
4 0

Answer:

your answer is

26 cm

hope it is helpful to you

allsm [11]3 years ago
3 0

Answer:

the diameters of circle with radius 13cm

  • 13×2
  • 26cm

hope it helps.

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A square is cut from a rectangle. The side length of the square is half of the unknown width w. The area of the shaded region is
Virty [35]

Answer:

  • See below

Step-by-step explanation:

<u>The side of the square is w/2, then its area is:</u>

  • A = (w/2)² = w²/4

<u>The shaded area is the difference of areas of the rectangle and the square:</u>

  • 14w - w²/4 = 84

<u>This can be simplified:</u>

  • 56w - w² = 336
  • w² - 56w + 336 = 0

8 0
2 years ago
Factor completely. 2x^4+4x^3-30x^2
kramer

Answer:

Step-by-step explanation:

2x⁴ + 4x³ - 30x² = 2x²*(x² + 2x - 15)

x² + 2x - 15

Sum = 2

Product = -15

Factors = 5 ; (-3)

x² + 2x - 15 = x² + 5x - 3x - 3 *5

                  = x(x + 5) - 3(x + 5)

                  = (x + 5)(x - 3)

2x⁴ + 4x³ - 30x² = (2x²) (x + 5)(x - 3)

7 0
3 years ago
X°<br> 1<br> 63°<br> m<br> n<br> find the value of x
Aleonysh [2.5K]

Answer:

I believe its 63°......,.,..,.,.,.,

8 0
3 years ago
There are 6 red marbles, 9 blue marbles, and 10 green marbles in a bag.
Nat2105 [25]

Answer:

I think 64%

Step-by-step explanation:

6+9+10 = 25

25x4 = 100

6x4 = 24

10x4 = 40

24+40 = 64

64/100 = 64%

Sry if this is wrong but I think this should be correct

7 0
3 years ago
Read 2 more answers
Solve (x-4)^2=5 get it correct and get brainliest
Marta_Voda [28]

Answer:

Two solutions were found :

x =(8-√20)/2=4-√ 5 = 1.764

x =(8+√20)/2=4+√ 5 = 6.236

Step-by-step explanation:

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    (x-4)^2-(5)=0

Step by step solution :

Step  1  :

1.1     Evaluate :  (x-4)2   =    x2-8x+16

Trying to factor by splitting the middle term

1.2     Factoring  x2-8x+11

The first term is,  x2  its coefficient is  1 .

The middle term is,  -8x  its coefficient is  -8 .

The last term, "the constant", is  +11

Step-1 : Multiply the coefficient of the first term by the constant   1 • 11 = 11

Step-2 : Find two factors of  11  whose sum equals the coefficient of the middle term, which is   -8 .

     -11    +    -1    =    -12

     -1    +    -11    =    -12

     1    +    11    =    12

     11    +    1    =    12

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  1  :

 x2 - 8x + 11  = 0

Step  2  :

Parabola, Finding the Vertex :

2.1      Find the Vertex of   y = x2-8x+11

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   4.0000  

Plugging into the parabola formula   4.0000  for  x  we can calculate the  y -coordinate :

 y = 1.0 * 4.00 * 4.00 - 8.0 * 4.00 + 11.0

or   y = -5.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2-8x+11

Axis of Symmetry (dashed)  {x}={ 4.00}

Vertex at  {x,y} = { 4.00,-5.00}

x -Intercepts (Roots) :

Root 1 at  {x,y} = { 1.76, 0.00}

Root 2 at  {x,y} = { 6.24, 0.00}

Solve Quadratic Equation by Completing The Square

2.2     Solving   x2-8x+11 = 0 by Completing The Square .

Subtract  11  from both side of the equation :

  x2-8x = -11

Now the clever bit: Take the coefficient of  x , which is  8 , divide by two, giving  4 , and finally square it giving  16

Add  16  to both sides of the equation :

 On the right hand side we have :

  -11  +  16    or,  (-11/1)+(16/1)

 The common denominator of the two fractions is  1   Adding  (-11/1)+(16/1)  gives  5/1

 So adding to both sides we finally get :

  x2-8x+16 = 5

Adding  16  has completed the left hand side into a perfect square :

  x2-8x+16  =

  (x-4) • (x-4)  =

 (x-4)2

Things which are equal to the same thing are also equal to one another. Since

  x2-8x+16 = 5 and

  x2-8x+16 = (x-4)2

then, according to the law of transitivity,

  (x-4)2 = 5

We'll refer to this Equation as  Eq. #2.2.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

  (x-4)2   is

  (x-4)2/2 =

 (x-4)1 =

  x-4

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:

  x-4 = √ 5

Add  4  to both sides to obtain:

  x = 4 + √ 5

Since a square root has two values, one positive and the other negative

  x2 - 8x + 11 = 0

  has two solutions:

 x = 4 + √ 5

  or

 x = 4 - √ 5

Solve Quadratic Equation using the Quadratic Formula

2.3     Solving    x2-8x+11 = 0 by the Quadratic Formula .

According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by            

           - B  ±  √ B2-4AC

 x =   ————————

                     2A

 In our case,  A   =     1

                     B   =    -8

                     C   =   11

Accordingly,  B2  -  4AC   =

                    64 - 44 =

                    20

Applying the quadratic formula :

              8 ± √ 20

  x  =    —————

                   2

Can  √ 20 be simplified ?

Yes!   The prime factorization of  20   is

  2•2•5

To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 20   =  √ 2•2•5   =

               ±  2 • √ 5

 √ 5   , rounded to 4 decimal digits, is   2.2361

So now we are looking at:

          x  =  ( 8 ± 2 •  2.236 ) / 2

Two real solutions:

x =(8+√20)/2=4+√ 5 = 6.236

or:

x =(8-√20)/2=4-√ 5 = 1.764

Two solutions were found :

x =(8-√20)/2=4-√ 5 = 1.764

x =(8+√20)/2=4+√ 5 = 6.236

Processing ends successfully

4 0
3 years ago
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