Can some one help me with this math problem?

Dividing the polynomial P(x) by x-6 yields a quotient by Q(x) and a remainder of 5.

bazaltina [42]

Dividing the given polynomial by (x -6) gives quotient Q(x) and remainder 5 then for Q(-6) = 3 , P(-6) = -31and P(6) =5.

As given in the question,

P(x) be the given polynomial

Dividing P(x) by divisor (x-6) we get,

Quotient = Q(x)

Remainder = 5

Relation between polynomial, divisor, quotient and remainder is given by :

P(x) = Q(x)(x-6) + 5 __(1)

Given Q(-6) = 3

Put x =-6 we get,

P(-6) = Q(-6)(-6-6) +5

⇒ P(-6) = 3(-12) +5

⇒ P(-6) =-36 +5

⇒ P(-6) = -31

Now x =6 in (1),

P(6) = Q(6)(6-6) +5

⇒ P(6) = Q(6)(0) +5

⇒ P(6) = 5

Therefore, dividing the given polynomial by (x -6) gives quotient Q(x) and remainder 5 then for Q(-6) = 3 , P(-6) = -31and P(6) =5.

The complete question is:

Dividing the polynomial P(x) by x - 6 yields a quotient Q(x) and a remainder of 5. If Q(-6) = 3, find P(-6) and P(6).

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7 0

1 year ago

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

WILL PICK BRAINLIEST!!!

andreyandreev [35.5K]

we are given

$f(x)=10(2)^x$

now, we can compare it with

$f(x)=a(b)^x$

we can find b

we get

$b=2$

now, we are given

How would the graph change if the b value in the equation is decreased but remains greater than 1

Let's take

b=1.8

$f(x)=10(1.8)^x$

b=1.6

$f(x)=10(1.6)^x$

b=1.4

$f(x)=10(1.4)^x$

b=1.2

$f(x)=10(1.2)^x$

now, we can draw graph

now, we will verify each options

option-A:

we know that all y-value will begin at y=0

because horizontal asymptote is y=0

so, this is FALSE

option-B:

we can see that

curve is moving upward when b decreases for negative value of x

but it is increasing slowly for negative values of x

so, this is FALSE

option-C:

we can see that

curve is moving upward when b decreases for negative value of x

but it is increasing slowly for negative values of x

so, this is TRUE

option-D:

we know that curves are increasing

so, the value of y will keep increasing as x increases

so, this is TRUE

option-E:

we can see that

curve is moving upward when b decreases for negative value of x

but it is increasing slowly for negative values of x

so, this is FALSE

4 0

3 years ago

Read 2 more answers

Can someone help me with these 2 questions please and thank you?

Inessa05 [86]

2 × 3 = 6. 2. 7×3 =21

4 0

3 years ago

The nine-digit number that identifies the bank that a check came from is called

Neko [114]

Answer: A

Step-by-step explanation:

7 0

3 years ago