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3 years ago

10

Solve the equations belowa. 6+10x = x+12

2 answers:

SOVA2 [1]3 years ago

7 0

Answer:0.6

Step-by-step explanation:

attashe74 [19]3 years ago

7 0

6+10x=x+12 the answer to that is 0.6

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Solve the equation show steps -2m+4m+5=3

Dafna11 [192]

-2m+4m+5=3
(-2m+4m)+(5)=3
2m+5=3
2m+5-5=3-5
2m=-2
2m/2=-2/2
m=-1

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mr Goodwill [35]

X=4
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Explain how 2 plus 2 Equals 4

aivan3 [116]

You have 2+2 in each 2 there are 1's so

2=(1+1) 2=(1+1) =4
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What is the GCF of 36 and 84

FinnZ [79.3K]

The greatest common factor( GCF) of 36 and 84 is 12.

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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