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3 years ago

-3x+18=7x what could you do to isolate the variable term to one side of the equation

Mathematics

2 answers:

Alexxandr [17]3 years ago

7 0

Answer:

7x=−x+24 7 x = − x + 24

Step-by-step explanation:

The equations we solved in the last section simplified nicely so that we could use the. Our strategy will involve choosing one side of the equation to be the variable side, and step by step, to isolate the variable terms on one side of the equation

BigorU [14]3 years ago

7 0

Answer:

We should add 3x to both sides, after that,then divide both sides of your result by 10

X=1.8

Step-by-step explanation:

-3x+18=7x

Add 3x to both sides

3x-3x+18=3x+7x

18=10x

Divide both sides by 10

18/10=10x/10

1.8=x

x=1.8

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x=-11x+4

x+11x=-11x+4+11x

12x=4

x= 4/12

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3 years ago

According to a survey, 15% of city workers take the bus to work. Donatella randomly surveys 10 workers. What is the probability

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Answer:0.001

x=the number of workers taking the bus to work

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3 years ago

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What is the mixed number for 18/15

bearhunter [10]

15 goes into 18 once with 3 remainder.

18/15-15/15=3/15=1/5

So the answer is:
$1 \frac{1}{5}$

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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

Solve:<br>3 = -2v - V<br>V=

Y_Kistochka [10]

Answer:

V = −1

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

3=−2v−v

3=−2v+−v

3=(−2v+−v)(Combine Like Terms)

3=−3v

Step 2: Flip the equation.

−3v=3

Step 3: Divide both sides by -3.

-3 -3

v=−1

4 0

3 years ago

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