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3 years ago

Simplify -2x^3y+xy^2

Mathematics

1 answer:

zheka24 [161]3 years ago

8 0

Answer:

-2x^4y^3 this is right. give me brainliest.

Step-by-step explanation:

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On his outward journey, Ali travelled at a speed of s km/h for 2.5 hours. On his return journey, he increased his speed by 4 km/

den301095 [7]

Answer:

3.08 km/h.

Step-by-step explanation:

We know that,

$Speed=\dfrac{Distance}{Time}$

Ali traveled at a speed of s km/h for 2.5 hours.

Let d be the distance and t be the time.

$2.5=\dfrac{d}{t}$

$2.5t=d$ ...(1)

On his return journey, he increased his speed by 4 km/h and saved 15 minutes. So, distance is d and times is t-15.

$4=\dfrac{d}{t-15}$

$4(t-15)=d$ ...(2)

From (1) and (2), we get

$4(t-15)=2.5t$

$4t-60-2.5t=0$

$1.5t=60$

$t=40$

Put t=40 in (1).

$d=2.5(40)=100$

So, t=40 and d=100.

Now,

Total distance = d + d = 100 + 100 = 200

Total time = t + t - 15 = 40 + 40 - 15 = 65

So, the average speed is

$\text{Average speed}=\dfrac{\text{Total distance}}{\text{Total time}}$

$\text{Average speed}=\dfrac{200}{65}$

$\text{Average speed}\approx 3.08$

Therefore, the average speed is 3.08 km/h.

3 0

4 years ago

If you have $100 and then lose $25 what is the percent change

katovenus [111]

$\huge{\textbf{\textsf{{\color{pink}{An}}{\red{sw}}{\orange{er}} {\color{yellow}{:}}}}}$

The lost percentage is 25%

The percentage remaining is 75%

Thanks
ThanksHope it helps

7 0

3 years ago

The height of a cone is twice the radius of its base.

fredd [130]

Answer: 2/3πx³

Step-by-step explanation:

Let the radius of the cone be represented by x.

Since the height of the cone is twice the radius of its base, the height will be: = 2x

Volume of a cone = 1/3πr²h

where,

r = x

h = 2x

Volume of a cone = 1/3πr²h

= 1/3 × π × x² × 2x

= 1/3 × π × 2x³

= 2/3πx³

Therefore, the correct answer is 2/3πx³.

3 0

3 years ago

For every arm a person has, they have a leg. Which ratio represents this situation?

Irina-Kira [14]

A: 1:1 unless they want you to count both arms then it would be B: 2:2

3 0

3 years ago

Read 2 more answers

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