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2 years ago

Find the like terms in the Algebraic Expression ( 2x +5x+10y+7)

Mathematics

1 answer:

Aliun [14]2 years ago

7 0

Answer:

The answer is B. 2x , 5x.

Step-by-step explanation:

This is the answer because like terms are alike. For example: With 2x and 5x what do they have in common? They have the x in common. Another example is if you have (2 + 7 + 5x + 10y). What are the like terms? 2 and 7, because they are both integers. Also, if you are trying to solve an expression such as the last example, then you would add 2 + 7, because they are similar, so they can be added. So it would be (9 + 5x + 10y).

Hope this helps! :)

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To estimate the number of lizards in a certain region, Sean traps 36 lizards and marks them. He releases these lizards and then

laiz [17]

Answer:

475

Step-by-step explanation:

i divided the two numbers and got 556 so your estimate should be 475

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2 years ago

Read 2 more answers

-7q + 12r = 3q - 4r what dose r equal

Elanso [62]

Answer:

r = 5y/8

Step-by-step explanation:

isolate the variable by dividing each side by factors that contain the variable.

6 0

2 years ago

3x+2y=-15 2x=-18 elimination

Sergeeva-Olga [200]

Answer: (-9,6)

Step-by-step explanation:

3x+2y=-15

2x=-18

First solve for x in second equation

2x=-18

2x/2=-18/2

X= -9

Now substitute your x into first equation to find y

3(-9)+2y= -15

-27+2y=-15

2y= -15+27

2y= 12

2y/2=12/2

Y= 6

5 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

What is the answer to 11/12×6=

Mademuasel [1]

Answer:

$\frac{11}{2}$

Step-by-step explanation:

$\frac{11}{12}$ x 6 = $\frac{11}{2}$

5 0

2 years ago