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Ad libitum [116K]

3 years ago

14

Please help asap !!!!!! Which equation has one solution for x?

2 answers:

White raven [17]3 years ago

5 0

Answer:

I think C

Step-by-step explanation:

A has no solution

B has no X

D has infinite solutions

steposvetlana [31]3 years ago

4 0

Question number b is the correct one

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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

What is 21 divided by 2 + 1 times 5 mines 2 to the second power?

goblinko [34]

21÷(2+1)5-2²
21÷10+5-4
21÷15-4
21÷11
21/11
1whole10divided by 11

5 0

4 years ago

Evaluate -3(9+y) if y = -7

mario62 [17]

Answer:

-6

Step-by-step explanation:

Substitutet -7 for y in -3(9+y): -3(9 - 7) = -3(2) = -6

6 0

4 years ago

Read 2 more answers

What are 5 irrational numbers

victus00 [196]

Answer:

square roots of:

2, 5, 3, 10, 15

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3 years ago

Rectangle ABCD is translated and then reflected to create rectangle A'B'C'D'. Do rectangle ABCD and rectangle A'B'C'D' have the

lions [1.4K]

Answer:

is there a graph for this that I can see?

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3 years ago