<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%20%2B%204%7D%7B3%7D%20%20%3D%206" id="TexFormula1" title=" \frac{x + 4}{3} = 6

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melisa1 [442]

4 years ago

" alt=" \frac{x + 4}{3} = 6" align="absmiddle" class="latex-formula">
how do u solve this

Mathematics

1 answer:

Gwar [14]4 years ago

8 0

Multiply both sides by 3 to cancel out the 3 in the denominator of the left side of the equal sign. You get:

x + 4 = 18

Now subtract 4 from both sides to get variables one side, constants on the other.

x = 18 - 4
x = 14

Your answer is x = 14

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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

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Semmy [17]

The answer :

4/5

Step-by-step explanation:

4x-5y=0

subtract 4x on both the sides

4x-4x-5y = 0-4x

-5y = -4x

divide by -5 on both the sides

-5y/-5 = -4x/-5

y =( 4/5 )x

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