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3 years ago

Pls help i give u brainliest amd 5 stars

Mathematics

1 answer:

MAVERICK [17]3 years ago

7 0

Answer:

160

Step-by-step explanation:

60-24=36 left

24 x 8 = 192

36/2=18

9 x 12 =108

108+192=300

300-140= 160 euros profit

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What function was used to make that pattern

Semenov [28]

Answer:

Step-by-step explanation:

The differences in the terms of f(x) are + 3, + 5, + 7

Since the differences are not constant the relationship is not linear

Note the differences in the differences are + 2, + 2,

The second differences are constant indicating a quadratic relationship

Note the relationship between x and f(x)

x = 1 → 1² = 1 ← require to add 5, that is 1 + 5 = 6 ← value of f(x)

x = 2 → 2² = 4 ← require to add 5, that is 4 + 5 = 9 ← value of f(x)

x = 3 → 3² = 9 ← require to add 5, that is 9 + 5 = 14 ← value of f(x)

x = 4 → 4² = 16 ← require to add 5, that is 16 + 5 = 21 ← value of f(x)

Thus f(x) = x² + 5 → B

3 0

3 years ago

Describe the end behavior of the given function <br> f(x)=9x^3-x^2-8x+6<br>

Mekhanik [1.2K]

As x approaches -inf f(x) -> -inf
and as x approaches inf, f(x) approaches +inf

Mark brainliest please

6 0

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).

4 0

3 years ago

16e^2-88e+121<br> pls help me

Rufina [12.5K]

Step-by-step explanation:-8.8e +122

7 0

4 years ago

Fill in the blank 3(2+7a)=6+

Lilit [14]

Answer:

$3(2+7a)=6+$