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3 years ago

5

How do you solve 4 [ 5 ( 12 + 3 ) - 2 ] - 7

1 answer:

laila [671]3 years ago

3 0

4 [ 5 (12 + 3) - 2 ] - 7

4 [ 5 (15) - 2 ] - 7

4 [ 75 - 2 ] - 7

4 [ 73 ] - 7

292 - 7

285

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The table shows the cost of downloading songs.

Sonbull [250]

Answer:

Step-by-step explanation:

rate of change=cost per song

so the first option

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3 years ago

What is the domain ??

prohojiy [21]

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[-8, 6]

Step-by-step explanation:

Domain is the x values. Look at the smallet x value, in this case -8, and the largest, in this case 6, and write them like a point using [] not (). You use [ or ] for numbers and ( or ) for infinity signs.

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2 years ago

If angle∠ A and angle∠ B are supplementary angles and angle∠ A is eighteight times as large as angle∠ B, find the measures of a

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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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