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2 years ago

What is 90 words in 2 minutes as a unit rate

Mathematics

2 answers:

Dmitriy789 [7]2 years ago

6 0

Answer:

<em>Hello, Jaesuk Sakai Here!! (^^</em>

Step-by-step explanation:

Words in 2 minutes = 90

Words in a minute = 90 / 2

= 45

So rate is 45 words/minutes

<em>Happy to Help~ jaesuk sakai!</em>

Gemiola [76]2 years ago

3 0

Answer:

Levi saved 3/5 of the amount he needs to buy a 75 dollar video game. he earns 7.50 per hour working at the pizza shopping town how many hours will he need to work to pay for the rest of the game show all of your work to solve this problem explain the Steps you used! thank you 6th grade form please

Step-by-step explanation:

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The sum of two polynomials is 8d5 – 3c3d2 + 5c2d3 – 4cd4 + 9. If one addend is 2d5 – c3d2 + 8cd4 + 1, what is the other addend?

Anit [1.1K]

Answer: The correct option is A.

Step-by-step explanation: We are given a polynomial which is a sum of other 2 polynomials.

We are given the resultant polynomial which is : $8d^5-3c^3d^2+5c^2d^3-4cd^4+9$

One of the polynomial which are added up is : $2d^5-c^3d^2+8cd^4+1$

Let the other polynomial be 'x'

According to the question:

$8d^5-3c^3d^2+5c^2d^3-4cd^4+9=x+(2d^5-c^3d^2+8cd^4+1)$

$x=8d^5-3c^3d^2+5c^2d^3-4cd^4+9-(2d^5-c^3d^2+8cd^4+1)$

Solving the like terms in above equation we get:

$x=(8d^5-2d^5)+(-3c^3d^2+c^3d^2)+(5c^2d^3)+(-4cd^4-8cd^4)+(9-1)$

$x=6d^5-2c^3d^2+5c^2d^3-12cd^4+8$

Hence, the correct option is A.

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3 years ago

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Is there any slope in this?

Angelina_Jolie [31]

Nope anyway marry Christmas

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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).

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3 years ago

What are the next two numbers? 0.3 , 0.7 , 1.1 , 1.5 ,

Dmitry [639]

There is an increment of 0.4 in each term. Next will be
[1.5+0.4], [1.5+0.4+0.4]
1.9, 2.3

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3 years ago

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An observer is standing in a lighthouse 96 feet above the level of the water. The angle of depression of a buoy is 22. What is t

lorasvet [3.4K]

Tan(theta)=opp/adj
Tan(22)=96/x
X=96/tan(22)

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3 years ago

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