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3 years ago

Someone please help quickly and fast please! For brainliest

Mathematics

1 answer:

ohaa [14]3 years ago

7 0

Answer:

Perimeter = 16 feet

Total Surface Area = 24 sq ft.

Step-by-step explanation:

Let's assume that each square is equal to 1 foot. Using this we can estimate the perimeter and area as the following. For perimeter, we simply add all the squares distances surrounding the shaded object.

Perimeter = 16 feet (4 feet for the top and bottom and 2 feet every diagonal)

For the area, we need to calculate the area of the square in the middle and then the area of the two triangles.

Square = L * W

Square = 4 * 4

Square = 16 sq ft

Triangle = 0.5 * Base * Height

Triangle = 0.5 * 4 * 2

Triangle = 4 sq ft

Both triangles = 4 sq ft * 2 = 8 sq ft.

Now we add both area totals:

Total Surface Area = 16 + 8

Total Surface Area = 24 sq ft.

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Volume= Help me please thanks so much

Artemon [7]

Answer:

30.7

Step-by-step explanation:

It's 30.7 because 707/23

4 0

2 years ago

Read 2 more answers

169/13 draw a quick picture with base ten blocks explain

Kitty [74]

I wish i could show u how to work it out but just do it or work it out the same way as the fractional numbers.

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4 years ago

Moby sold half his toy car collection, and then bought 12 more cars the next day. He now has 48 toy cars. How many cars did he h

castortr0y [4]

Answer: 72 cars.

Step-by-step explanation:

Let be "x" the number of toys that Moby had to start with.

According to the information given in the exercise, Moby sold half his toy car collection. This can be represented with the following expression:

$\frac{1}{2}x$

Then he bought 12 more cars, giving a total amount of 48 toy cars.

Therefore, the equation that represents this situation is the equation shown below:

$x-\frac{1}{2}x+12=48$

Now you must solve for "x" in order to find its value.

You get that this is:

$\frac{1}{2}x+12=48\\\\\frac{1}{2}x=48-12\\\\\frac{1}{2}x=36\\\\x=(36)(2)\\\\x=72$

7 0

4 years ago

What is the definition of congruent line segments

Kay [80]

Are simply line segments that are equal in length. Congruent means equal.Congruent line segments are usually indicated by drawing the same amount of little tic lines in the middle of the segments, perpendicular to the segments. We indicate a line segment by drawing aline over its two endpoints.

7 0

3 years ago

Read 2 more answers

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