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2 years ago

Which equation shows y=−12x+4y=−12x+4 in standard form

Mathematics

1 answer:

astraxan [27]2 years ago

3 0

It cant be made into a standard form

Sorry if incorrect

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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

Please help pic attatched i will give brainliest

Alchen [17]

Answer:

N'(x)=90 In(43)*43^x-90 In(50)*43^x /50^x

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Max is painting a fence.The table shows the number of square feet he painted after various numbers of minutes. What is the rate

ohaa [14]

Answer:

3.5

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Find the integer values of x that satisfy both of these inequalities.

Shtirlitz [24]

Answer:

0, 1, 2.

Step-by-step explanation:

5 - 3x <= 7

-3x <= 2

x >= -2/3 (the inequality sign flips as we are dividing by a negative value)

4x + 1 < 13

4x < 12

x < 3.

So the integer values satisfying the inequalities are>

0, 1, 2.

6 0

2 years ago

Read 2 more answers

Anyone know the answer ?

LenaWriter [7]

Answer:

102

Step-by-step explanation:

We can find x by using the fact that the angles in a triangle add up to 180.

73+42+x=180

x = 65

We can find y by using the fact that angles on a straight line add up to 180.

131+y=180

y=49

We can find the third angle in that triangle.

65+49+n=180

Let n represent the third angle in the triangle.

n=66

Then z + n = 180

z + 66 =180

z= 144

Let the angle in the bottom left be i

z+i = 180

144 + i = 180

i = 36

Let's find l.

The angles in a triangle add up to 180.

36 + 49 + l =180

l = 95

y is a vertical angle.

95+ y + u = 180

95+49+u=180

u=36

180-36=144

then the right triangle (with angle of 42) has a corner of 36.

36+42+t = 180

t= 102

102 + f = 180

f = 78

f + w = 180

78 + w = 180

w = 102

3 0

3 years ago