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riadik2000 [5.3K]

3 years ago

13

HELP HELP HELP PLEASE ASSP

2 answers:

Fynjy0 [20]3 years ago

4 0

Answer:

2nd option

Step-by-step explanation: if the numbers keeps on repeating in the same pattern after a decimal point then they're considered to be non terminating and repeating and rational

Diano4ka-milaya [45]3 years ago

3 0

(i) is the rational and (ii) is the rational.

Is your answer. Hope this helps

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Linear pair (x-39) + (x+61)

faust18 [17]

(x - 39) + (x + 61) = x - 39 + x + 61 = (x + x) + (-39 + 61) = 2x + 22

7 0

4 years 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

Just answer this question for points

Kobotan [32]

Answer:

Thanks You the BEST.

Step-by-step explanation:

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Hello please help me with this question

Sunny_sXe [5.5K]

A: 2/1 -> 2
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Y value is up and down and x value is side to side.
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3 years ago

Many graduates earn degrees in fields that result in high-paying jobs. Based on that assumption, which two fields most likely pa

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