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2 years ago

Find the equation of the circumference of a wheel if the radius is 2.5.

Mathematics

1 answer:

dimaraw [331]2 years ago

5 0

Answer:it is d

Step-by-step explanation:

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explain how estimated the quotient helps you place the first digit in the quotient of a division problems

Marysya12 [62]

If i understood it right when you estimate the quotient it help you have an idea

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2 years ago

The grade of a road is the ratio of its rise to its run and is usually given as a decimal or percent. Find the angle that this r

Andru [333]

Answer:

the angle would be 1/2 i did this problem before and got it correct

Step-by-step explanation:

3 0

2 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

2 years ago

Help ASAP!! Geometry -

horrorfan [7]

Answer:

x = 9.27 cm

Volume = 289.2 cm³

Step-by-step explanation:

SA = Surface Area

SA = 229.2 cm²

P = Perimeter of Base

P = 3(6)

P = 18 cm

H = Height of Solid

H = x

A = Area of Base(Equilateral Triangle)

A = 6²√3 / 4

A = 36√3 / 4

A = 9√3

A = 9(1.732)

A = 31.176

A = 31.2 cm²

SA = Surface Area

SA = PH + 2A

229.2 = 18x + 2(31.2)

229.2 = 18x + 62.4

229.2 - 62.4 = 18x

166.8 = 18x

9.266 = x

9.27 = x

x = 9.27 cm

V = Volume

V = AH

V = 31.2(9.27)

V = 289.224

V = 289.2 cm³

8 0

2 years ago

PLEASE HELP ME FOR EXTRA POINTS AND BRAINLIEST ANWSER

NNADVOKAT [17]

Recall that tanA = sinA / cosA.

Therefore, tanA = (3/5) / (4/5) = 3/4.

3 0

3 years ago