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Marina86 [1]
3 years ago
6

Which triangle always has only two sides the same length and one angle that measures 90°? A. acute scalene B. right isosceles C.

obtuse scalene D. right scalene
Mathematics
1 answer:
Otrada [13]3 years ago
5 0
Right isosceles, which has two sides the same length and one angle that measures 90 degrees.
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The government plans to build 75,000 new homes. 1/5 of the new homes will be built in a city near you. How many is that?
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4 0
3 years ago
Find the length of a rectangle if its
Alina [70]

Answer: L = 29cm

Step-by-step explanation:

Let's go!!

To find any rectangle area, you have to apply this formula: A = L . w , where L is the length and w is the width.

So, we have w = L - 13 (the width is 13 less than the Length) and then, perimeter is 90 cm.

Well, I hope you remember that the perimeter of rectangle is 2L + 2W. In this case, 2L + 2w = 90

Then, you might to solve this system of equation:

2L + 2w = 90

w = L - 13

Simplifying the first equation, you'll have L + w = 45 (you can divide everything of 2).

Our new system:

L + w = 45

w = L - 13

Using the substituition method:

L + L - 13 = 45

2L = 58

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4 0
3 years ago
Which of the following statements about the polynomial function f(x)=x^3+2x^2-1
ch4aika [34]

x = -1

x =(1-√5)/-2= 0.618

x =(1+√5)/-2=-1.618

Step  1  :

Equation at the end of step  1  :

 0 -  (((x3) +  2x2) -  1)  = 0  

Step  2  :  

Step  3  :

Pulling out like terms :

3.1     Pull out like factors :

  -x3 - 2x2 + 1  =   -1 • (x3 + 2x2 - 1)  

3.2    Find roots (zeroes) of :       F(x) = x3 + 2x2 - 1

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  -1.

The factor(s) are:

of the Leading Coefficient :  1

of the Trailing Constant :  1

Let us test ....

  P    Q    P/Q    F(P/Q)     Divisor

     -1       1        -1.00        0.00      x + 1  

     1       1        1.00        2.00      

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that

  x3 + 2x2 - 1  

can be divided with  x + 1  

Polynomial Long Division :

3.3    Polynomial Long Division

Dividing :  x3 + 2x2 - 1  

                             ("Dividend")

By         :    x + 1    ("Divisor")

dividend     x3  +  2x2      -  1  

- divisor  * x2     x3  +  x2          

remainder         x2      -  1  

- divisor  * x1         x2  +  x      

remainder          -  x  -  1  

- divisor  * -x0          -  x  -  1  

remainder                0

Quotient :  x2+x-1  Remainder:  0  

Trying to factor by splitting the middle term

3.4     Factoring  x2+x-1  

The first term is,  x2  its coefficient is  1 .

The middle term is,  +x  its coefficient is  1 .

The last term, "the constant", is  -1  

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

Step-2 : Find two factors of  -1  whose sum equals the coefficient of the middle term, which is   1 .

     -1    +    1    =    0  

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  3  :

 (-x2 - x + 1) • (x + 1)  = 0  

Step  4  :

Theory - Roots of a product :

4.1    A product of several terms equals zero.  

When a product of two or more terms equals zero, then at least one of the terms must be zero.  

We shall now solve each term = 0 separately  

In other words, we are going to solve as many equations as there are terms in the product  

Any solution of term = 0 solves product = 0 as well.

Parabola, Finding the Vertex :

4.2      Find the Vertex of   y = -x2-x+1

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -0.5000  

Plugging into the parabola formula  -0.5000  for  x  we can calculate the  y -coordinate :  

 y = -1.0 * -0.50 * -0.50 - 1.0 * -0.50 + 1.0

or   y = 1.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -x2-x+1

Axis of Symmetry (dashed)  {x}={-0.50}  

Vertex at  {x,y} = {-0.50, 1.25}  

x -Intercepts (Roots) :

Root 1 at  {x,y} = { 0.62, 0.00}  

Root 2 at  {x,y} = {-1.62, 0.00}  

Solve Quadratic Equation by Completing The Square

4.3     Solving   -x2-x+1 = 0 by Completing The Square .

Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:

x2+x-1 = 0  Add  1  to both side of the equation :

  x2+x = 1

Now the clever bit: Take the coefficient of  x , which is  1 , divide by two, giving  1/2 , and finally square it giving  1/4  

Add  1/4  to both sides of the equation :

 On the right hand side we have :

  1  +  1/4    or,  (1/1)+(1/4)  

 The common denominator of the two fractions is  4   Adding  (4/4)+(1/4)  gives  5/4  

 So adding to both sides we finally get :

  x2+x+(1/4) = 5/4

Adding  1/4  has completed the left hand side into a perfect square :

  x2+x+(1/4)  =

  (x+(1/2)) • (x+(1/2))  =

 (x+(1/2))2

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

  x2+x+(1/4) = 5/4 and

  x2+x+(1/4) = (x+(1/2))2

then, according to the law of transitivity,

  (x+(1/2))2 = 5/4

We'll refer to this Equation as  Eq. #4.3.1  

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

Note that the square root of

  (x+(1/2))2   is

  (x+(1/2))2/2 =

 (x+(1/2))1 =

  x+(1/2)

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:

  x+(1/2) = √ 5/4

Subtract  1/2  from both sides to obtain:

  x = -1/2 + √ 5/4

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

  x2 + x - 1 = 0

  has two solutions:

 x = -1/2 + √ 5/4

  or

 x = -1/2 - √ 5/4

Note that  √ 5/4 can be written as

 √ 5  / √ 4   which is √ 5  / 2

Solve Quadratic Equation using the Quadratic Formula

4.4     Solving    -x2-x+1 = 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   =    -1

                     C   =   1

Accordingly,  B2  -  4AC   =

                    1 - (-4) =

                    5

Applying the quadratic formula :

              1 ± √ 5

  x  =    ————

                  -2

 √ 5   , rounded to 4 decimal digits, is   2.2361

So now we are looking at:

          x  =  ( 1 ±  2.236 ) / -2

Two real solutions:

x =(1+√5)/-2=-1.618

or:

x =(1-√5)/-2= 0.618

Solving a Single Variable Equation :

4.5      Solve  :    x+1 = 0  

Subtract  1  from both sides of the equation :  

                     x = -1

Hope this helps.

6 0
3 years ago
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