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3 years ago

2+2355433455333338387463783837747383883773738

Mathematics

2 answers:

nordsb [41]3 years ago

5 0

The calculator says it’s 2.355433455333e42

Step-by-step explanation:

Ivahew [28]3 years ago

3 0

Answer:

2.3554335e+42

Step-by-step explanation:

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Select the number of punctuation error. Immediately –1 even before I could stop the water,2 another leak occurred,3 and water gu

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Step-by-step explanation:

Select the number of punctuation error. Immediately even before I could stop the water, another leak occurred, and water gushed onto the, kitchen floor. A. 1 B. 2 C. 3 D. 4

3. There doesn't need to be a comma after 'the'.

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What value of y makes the equation true. y+2.9=11

yawa3891 [41]

Answer:

y = 8.1

Step-by-step explanation:

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3 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

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

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

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

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

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

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

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

$N(1)=$

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

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

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

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

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

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3 years ago

Which inequality is NOT true when x= -2?

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Answer: D

Step-by-step explanation:

-7 isn't greater than -4

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2 years ago