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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Mathematics

1 answer:

gavmur [86]3 years ago

6 0

$\text{Use}\\\\a^2-b^2=(a-b)(a+b)\\\\i=\sqrt{-1}\to i^2=-1\\---------------------\\\\x^2+25=x^2+5^2=x^2-(-1)(5^2)=x^2-(i^2)(5^2)=x^2-(5i)^2\\\\\boxed{x^2+25=(x-5i)(x+5i)}$

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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

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

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

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

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

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

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

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

$N(1)=$

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

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

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

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

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

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Answer:

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One angle is inside a right triangle

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3 years ago

Read 2 more answers

2) A fair coin was tossed 75 times. The coin landed on Heads 42 times and Tails 33 times. What is the experimental probability o

coldgirl [10]

Answer:

$\frac{11}{25}$

Step-by-step explanation:

In the experiment that was conducted the coin was tossed a total of 75 times and out of those times it only landed on tails 33 times. Therefore the experimental probability of the coin landing on tails can be calculated by the dividing the times it landed on tails by the total number of times it was tossed. Like so...

$\frac{33}{75}$ or 0.44