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3 years ago

I can’t seem to find any answers for this can anyone help?

Mathematics

1 answer:

ElenaW [278]3 years ago

7 0

Answer:

2 2/3

Step-by-step explanation:

- - is the same as + so it's actually 2 1/3 + 5

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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

2 (y+6) + 6 (y+6) Expand and simplify

san4es73 [151]

Hi there! The answer is 8y + 48.

$2(y + 6) + 6(y + 6)$
First work out the parenthesis.

$2y + 12 + 6y + 36$
Rewrite the expression in order to collect the terms.

$2y + 6y + 12 + 36$
Next up, collect the terms.

$8y + 48$
Now we've found our answer.

5 0

3 years ago

Two chords in a circle intersect. One chord is made of 6 and 5, and the other is made of x +1 and x. What is x?

Zolol [24]

Answer:

x = 5 and -6

Step-by-step explanation:

Using the intersecting chord theorem which states that the products of the lengths of the line segments on each chord are equal.

Hence:

let

a = 6, b = 5, c = x+1 and d = x

Therefore, ab = cd