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3 years ago

Determine g(-1) for g(t)=(t^2-1)(t+4) A)-2 B)-6 C)3 D)0

2 answers:

7 0

It is C because Since {{g(x)}={x^3+2}}g(x)=x3+2g, left parenthesis, x, right parenthesis, equals, x, start superscript, 3, end superscript, plus, 2, we can substitute {x^3+2}x3+2x, start superscript, 3, end superscript, plus, 2 in for {g(x)}g(x)g, left parenthesis, x, right parenthesis.\begin{aligned}{f(g(x))}&=3(g(x))-1 \\\\ &=3({x^3+2})-1 \\\\ &=3x^3+6-1\\\\ &=3x^3+5 \end{aligned}f(g(x))=3(g(x))−1=3(x3+2)−1=3x3+6−1=3x3+5

Zanzabum3 years ago

3 0

D option that is 0 will be answer

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Please help. !!!!! Only if you are good at college algebra

kicyunya [14]

Yes I got an A in college algebra...

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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).

4 0

3 years ago

The prices of tickets for the local rock concert were $16 for Friday night and $28 for Saturday night. The total attendance for

ASHA 777 [7]

Let x: number of tickets that sold on Friday, y: number of tickets that sold on Saturday night

16x+28y=24232
x+y=1090
x=1090-y,
so 16(1090-y)+28y=24232
17440-16y+28y=24232
12y=6792
y=566. As a result, there will be 566 tickets sold on Saturday night. Hope it help!

4 0

3 years ago

Read 2 more answers

Will mark brainlist if right 15 pts

aleksley [76]

Answer:

$g(x)=x-1\\g(f(x))=(-2x-5)-1\\g(f(x)=-2x-6$

Step-by-step explanation:

So this is a composition of functions. Think back to solutions of equations. You would have to put the equation into the other equations. It's like this!

So I graphed it like this on desmos:

$g(x)=x-1\\g(f(x))=(-2x-5)-1\\g(f(x)=-2x-6$

3 0

2 years ago

A bag has 3 red marbles, 3 blue, 4 green and 2 yellow. What is the theoretical probability of pulling a red marble?

svetlana [45]

Answer:

Fraction: $\frac{1}{4}$

Decimal: 0.25

Percentage: 25%

Step-by-step explanation:

Have a good day :)

4 0

3 years ago