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3 years ago

Simplify 7n - 2n 3p 2p.

Mathematics

2 answers:

djyliett [7]3 years ago

8 0

5n+5p is the right answer.

yan [13]3 years ago

5 0

Answer:

5n+5p

Step-by-step explanation:

thats the answer. hope this helps!!! :) :p

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What’s the answer ??? ( someone please help )

Fofino [41]

It would probably be B because it said between -30 and 130

3 0

3 years ago

5. Ms. Lee has 220 students in all her classes. If her classes collected 2860 pounds of food, how many meals did each student pr

jarptica [38.1K]

Answer:

Each students provides 13 pounds of meal

Step-by-step explanation:

Total students in Ms. Lee class = 220

Total pounds of food collected = 2,860

how many meals did each student provide?

Meals provided by each student = Total pounds of food collected / total number of students

= 2,860 / 220

= 13 pounds

Each students provides 13 pounds of meal

Note: 1.2 pounds of food equals one meal

1.2 pounds : 1 meal = 13 pounds : x meals

1.2/1 = 13/x

1.2 * x = 1 * 13

1.2x = 13

x = 13 / 1.2

= 10.83 meals

Approximately 11 meals to the nearest number

Note:1 kilogram is 2.2 pounds

1 kg : 2.2 pounds = x kg : 13 pounds

1/2.2 = x / 13

1 * 13 = 2.2 * x

13 = 2.2x

x = 13/2.2

= 5.91 kilogram

Approximately x = 6 kilograms

3 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

3 years ago

What is the exact circumference of the circle?

tensa zangetsu [6.8K]

Your answer would be B.)

3 0

3 years ago

Read 2 more answers

What is angle A? 313 degrees 133 degrees 143 degrees 147 degrees

Evgesh-ka [11]

133 is the Answer bc that is a supplementary angle and supplementary angles always sum up to 180 hope this helps please mark brainliest

Step-by-step explanation:

5 0

3 years ago