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3 years ago

6

A three-toed sloth has a top speed of 0.22 feet per second. A giant tortoise has a top speed of 2.992 inches per second. Convert

both speeds to miles per hour, and determine which animal is faster.

1 answer:

Akimi4 [234]3 years ago

6 0

If you convert the sloth it is
0.15 miles per hour
If you convert the giant tortoise it is
0.17 miles per hour
You can then compare the two and realize the giant tortoise is 0.02 miles per hour faster than the sloth

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Write the standard form of the equation of line through (3, -1) with a slope of -1.

kirill115 [55]

The standard form of a line: $Ax+By=C$ .

The point-slope form: $y-y_1=m(x-x_1)$

We have the point (3, -1) and the slope m = -1. Substitute:

$y-(-1)=-1(x-3)$

$y+1=-x+3$ <em>add x to both sides</em>

$x+y+1=3$ <em>subtract 1 from both sides</em>

$\boxed{x+y=2}$

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3 years ago

Determine whether y varies with x. If so, find the constant of variation and write the function rule.

ira [324]

Answer: Yes, y does vary directly with x.

Constant of variation = 1/4

The function rule is y = (1/4)x

=========================================================

Explanation:

Let's assume that y does vary directly with x.

If that's the case, then we have an equation in the form y = kx, where k is the constant of variation.

Solving for k gets us k = y/x

For each row, divide the y value over the x value

row one: k = y/x = 14/56 = 0.25
row two: k = y/x = 20/80 = 0.25
row three: k = y/x = 22/88 = 0.25

Each row yields the value k = 0.25 and it fully confirms y does vary directly with x.

So y = kx becomes y = 0.25x as the function rule, which is equivalent to y = (1/4)x

6 0

1 year ago

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

const2013 [10]

Answer:

a.

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

b.

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

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

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

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

a.

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

b.

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

Mr.O'Brien is paid $10 per hour for the first 40 hours he works in a week. He is paid 1.5 times that rate for each hour after th

erma4kov [3.2K]

In first 40 hours he earns = 7.30 x 40 = $292
Each hour after his rate = 7.20 x 1.5 = $10.80 per hour
In extra four hours he earns = 10.80 x 4 = $43.20

Total earnings = $292 + $43.20 = $335.20
Therefore you don't agree
Hoped this helped you

8 0

3 years ago

(5 Points)<br> all (1 – 3K)<br> -396<br> 0-374<br> 0 -198<br> 0 -187

Triss [41]

Answer:−3K−395,−374,−198,−187

Step-by-step explanation:

Remove parentheses.

1−3K−396,0−374,0−198,0−187

Simplify 1-3K-3961−3K−396 to -3K-395−3K−395.

−3K−395,0−374,0−198,0−187

Simplify 0-3740−374 to -374−374.

−3K−395,−374,0−198,0−187

Simplify 0-1980−198 to -198−198.

−3K−395,−374,−198,0−187

Simplify 0-1870−187 to -187−187.

−3K−395,−374,−198,−187

6 0

2 years ago