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3 years ago

Identify and explain the error

1 answer:

3 0

You would add 10 to 50 then divide both sides of the equation by -10 and you would get x=4 as your final answer. I recommend using the app Photomath:)

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In a classroom of 33 students, the ratio of boys to girls is 3 : 8.

kirill [66]

Answer:

the answer is 9 :)

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Find the lateral area of the cylinder. Give your answer in terms of pi

Artist 52 [7]

Answer:

<h2>204π units²</h2>

Step-by-step explanation:

The lateral area of the cylinder includes both the side and the ends.

The area of the side can be found by calculating the circumference of the cylinder and multiplying that by the height: A = 2π(6 units )(11 units) = 132π units².

The area of one end of this cylinder can be found by applying the "area of a circle" formula: A = πr². Here, with r = 6 units, A = π(6 units)² = 36π units². Since the cylinder has two ends, the total area of the ends is thus 2(36π units) = 72π units.

The total lateral area of the cylinder is thus 72π units² + 132π units², or 204π units²

7 0

3 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

Please Help me A.S.A.P.

kati45 [8]

Answer:

16.01%

Step-by-step explanation:

45-28.99 is 16.01, so it's 16.01%.

4 0

2 years ago

!Help !

svetoff [14.1K]

Answer:

50 minutes

Step-by-step explanation:

she spent 1 hr and 30 minutes on homework

she spends 2/3 of an hour on the phone = 40 minutes

1 hr 30 min + 40 minutes = 2 hrs and 10 min

so she had 50 minutes to kill before bedtime since she goes to sleep 3 hrs after dinner

2hr 10 min + 50 min = 3 hrs.

8 0

3 years ago