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Montano1993 [528]
3 years ago
11

the sides of the rectangle are inreased by a scale factor of 4. The perimeter of the smaller retangle is 20 cm. What is the peri

meter of the larger rectangle?
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0
The height of the smaller rectangle, let us say is 5
5 x 4 = 20
(multiply height of rectangle to scale factor)
the length is 4
4 x 4 = 16
(multiply length of rectangle to scale factor)

to find larger factor, divide smaller with larger
16/20 = 1/4

scale factor should be 1/4

multiply 20 with 4

perimeter of larger should be 80

hope this helps

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Find the distance between the points (8,1) and (2,10).<br> Round decimals to the nearest tenth.
miss Akunina [59]

The nearest distance between the points (8,1) and (2,10) is 11 units, which is mentioned in and also calculated by using formula.

Step-by-step explanation:

The given is,

             Two points are (8,1) (2,10)

Step: 1

            By graphical method, refer the attachment,

                    First point (8,1)

                           The values of x=8 and y=1 noted in the graph

                    Second point (2,10)

                          The values of x=2 and y=10 noted in the graph

            Now join the two points (8,1) and (2,10)

                          Measure the distance between two points, the distance is 11 units. Due to some error the value becomes 10.98 or 11.1 we need to convert the answer into nearest whole number.

                                                     ( OR )

Step:1

        By formula method,

                   Distance =  \sqrt{(x_{2} - x_{1} )^{2} +(y_{2} - y_{1} )^{2} }.....................(1)

        Where,

                    (8,1) are (x_{1},y_{1})

                    (2,10) are (x_{2},y_{2})

       From the equation,

                                 = \sqrt{(2 - 8 )^{2} +(10 - 1 )^{2} }

                                 = \sqrt{(-6 )^{2} +(9 )^{2} }

                                 = \sqrt{36 +81 }

                                  = \sqrt{117}

                                  = 10.81665

                                  ≅ 11             ( Answer convert to the nearest tenth)

                  Distance = 11 units

Result:

               The nearest distance between the points (8,1) and (2,10) is 11 units, which is mentioned in and also calculated by using formula.

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3 years ago
Which equation can be used to determine the distance between the origin and (–2, –4)?
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3 years ago
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How do i solve for x ​
sergiy2304 [10]

Expplanation and Answer:

y=mx−7

Swap sides so that all variable terms are on the left hand side.

mx−7=y

Add 7 to both sides.

mx=y+7

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m

mx

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=

m

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Dividing by m undoes the multiplication by m.

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2 years ago
Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (
tekilochka [14]

Answer:

Radius of convergence of power series is \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n        n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n       n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

Power series centered at x = a is:

\sum_{n=1}^{\infty}c_{n}(x-a)^{n}

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]

Applying the ratio test:

\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}

\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

Applying n → ∞

\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

The numerator as well denominator of \frac{a_{n}}{a_{n+1}} are polynomials of fifth degree with leading coefficients:

(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}

4 0
3 years ago
The age of father is 4nyears.If the age of his daughter is(2n-9)years.find their total age.​
8_murik_8 [283]

Answer:

4n+(2n-9)

4n+2n-9

6n-9 =the total age

6n=9

divide by 6 both sides.

n=9/6

n=1.5 years.

substitute the value of n in the age of the father and the daughter.

Age of father=4×1.5

6years.

Age of Daughter = (2×1.5)-9

3-9

=-6 years

Total age

6-6=0

4 0
3 years ago
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