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Vaselesa [24]
3 years ago
11

Which expressions are equivalent to (2 Superscript 5 Baseline) Superscript negative 2? 2–10 and StartFraction 1 Over 20 EndFract

ion 2–10 and StartFraction 1 Over 1024 EndFraction 10–2 and StartFraction 1 Over 100 EndFraction 10–10 and StartFraction 1 Over 100 EndFraction
Mathematics
1 answer:
Serggg [28]3 years ago
4 0

Answer:

\frac{1}{2^{10}} \ and\ \frac{1}{1024}

Step-by-step explanation:

Given

(2^{5})^{-2}

Required

Find the equivalent;

To find the equivalent of the given expression, we make use of laws of indices;

Using the following law of indices;

(a^m)^n = a^{m*n}

So;

(2^{5})^{-2} becomes

(2^{5})^{-2} = 2^{5*-2}

(2^{5})^{-2} = 2^{-10} ------------ This is one equivalent

Solving further;

Using the following law of indices;

a^{-m} = \frac{1}{a^m}

So;

(2^{5})^{-2} = 2^{-10} becomes

(2^{5})^{-2} = \frac{1}{2^{10}}

2^10 = 1024

Hence;

(2^{5})^{-2} = \frac{1}{1024}

Conclusively; the equivalents of (2^{5})^{-2} are \frac{1}{2^{10}} \ and\ \frac{1}{1024}

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If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle
stira [4]

Answer:

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<u>Step-by-step explanation:</u>

Given points end  Points are p(-3,-2) and q( 1,6)

<em>The distance of two points formula</em>

P Q = \sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2}   }

P Q = \sqrt{1 - (-3)^{2} + ((6 -(-2))^{2}   }

P Q = \sqrt{16+64} = \sqrt{80}

<em>The diameter 'd' = 2 r</em>

                       2 r = √80

                            = \sqrt{16 X 5}

                           = 4 \sqrt{5}

                     <em> r =  2√5</em>

<em>Mid-point of two end points </em>

<em>                        </em>(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )<em></em>

<em>                                               = (-1 ,2)</em>

<em>Mid-point of two end points = center of the circle</em>

<em>                                     (h,k) = (-1 , 2)</em>

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

<em>          (x -(-1) )² +(y-(2))² = (2(√5))²</em>

<em>         x² + 2 x + 1 + y² - 4 y + 4 = 20</em>

<em>          x² + 2 x + y² - 4 y  = 20 -5</em>

<em>           x² + 2 x + y² - 4 y  = 15</em>

<u><em>Final answer</em></u><em>:-</em>

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<em>                  </em>

6 0
4 years ago
Please help me!!!!!!!!!!!​
harkovskaia [24]

Step-by-step explanation:

i think the answer is 10600

5 0
3 years ago
I cant get this one, 216−(−83)+(−479)
Inessa [10]

Answer:

180

Step-by-step explanation:

=216-(-83)+(-479)

=216+83-479

=299-479

=180

7 0
4 years ago
6-a/a+5 =
Anni [7]
<span>d) (6-a) divide (a+5) </span>
7 0
3 years ago
Write a polynomial f(x) that meets the given conditions. Answers may very
AnnZ [28]

The polynomial is: f(x)=x^3-2x^2-23x+60

Step-by-step explanation:

We need to write a polynomial f(x) that meets the given conditions: Degree 3 polynomial with zeros 3,-5,4

Since zeros are at 3,-5 and 4

So, x=3, x=-5 and x=4

or:

x-3=0, x+5=0 and x-4=0

Multiplying all factors to find the polynomial:

(x-3)(x+5)(x-4)=0\\(x(x+5)-3(x+5))(x-4)=0\\(x^2+5x-3x-15)(x-4)=0\\(x^2+2x-15)(x-4)=0\\x(x^2+2x-15)-4(x^2+2x-15)=0\\x^3+2x^2-15x-4x^2-8x+60=0\\x^3+2x^2-4x^2-15x-8x+60=0\\x^3-2x^2-23x+60=0

So, the polynomial is: f(x)=x^3-2x^2-23x+60

keywords: Factors and zeros of polynomial

Learn more about Factors and zeros of polynomial at:

  • brainly.com/question/2568692
  • brainly.com/question/1332667
  • brainly.com/question/1357167

#learnwithBrainly

4 0
3 years ago
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