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3 years ago

REALLY NEED HELP ON THESE 3 MATH QUESTIONS!!! (20 PTS)

Mathematics

1 answer:

hammer [34]3 years ago

7 0

1. x² - 6x + 8
4*2 equals 8 and if you add 4 and 2 it gives you 6
-4-2= -6
=(x-4)(x-2)

3. the answer is A
if you plug in the points into the equation y=x
y= 1/2x + 4
given point: (4,6) x=4 y=6
6= 1/2(4)+4
6=6

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Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

6 0

3 years ago

A newly hired basketball coach promised a high-paced attack that will put more points on the board than the team’s previously te

puteri [66]

Answer:

a. z = 2.00

Step-by-step explanation:

Hello!

The study variable is "Points per game of a high school team"

The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)

The hypothesis is:

H₀: μ ≤ 99

H₁: μ > 99

α: 0.01

There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.

The statistic value under the null hypothesis is:

Z= X[bar] - μ = 101 - 99 = 2

σ/√n 6/√36

I don't have σ, but since this is an approximation I can use the value of S instead.

I hope it helps!

7 0

3 years ago

15 POINTS WILL MARK BRAINLIEST!!!!! Choose the group responsible for the Declaration of Independence.

Bess [88]

Answer: second Congress

Step-by-step explanation:

Short and sweet

6 0

3 years ago

Find the values of x and y. The diagram is not to scale.

Aleksandr [31]

Step 1

Find the value of x

we know that

$(x-1)\°=74\°$ -------> by corresponding angles, because the given lines are parallel

Solve for x

$(x-1)\°=74\°\\x=74+1 \\ x=75\°$

Step 2

Find the value of y

we know that

$(x-1)+40+(y-2)=180$

substitute the value of x and solve for y

$(75-1)+40+(y-2)=180$

$74+40-2+y=180$

$112+y=180$

$y=180-112$

$y=68\°$

therefore

the answer is the option B

$x=75\°, y=68\°$

8 0

3 years ago

Read 2 more answers

Solve for s. 4s + 6s - 4 = 6

UkoKoshka [18]

Answer:

s = 1

Step-by-step explanation:

4s + 6s - 4 = 6

10s - 4 = 6

+ 4 +4

10s = 10

10s/10=

s = 1

Hope this helps! If you have any additional questions please don't hesitate to ask me or your teacher to be sure you master the subject. :) Stay safe and please mark brainliest!

6 0

3 years ago