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3 years ago

Help lots of points!!!

Mathematics

1 answer:

blondinia [14]3 years ago

8 0

Answer:

5/8

Step-by-step explanation:

would be reduces all the way by 3

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How many soulutions are there to the equation below? 8x+47=8(x+5)

Novosadov [1.4K]

Answer:

Step-by-step explanation:

8x+47=8(x+5)

8x+47=8x+40

47=40

which is impossible so it has no solution.

6 0

2 years ago

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

Find the value of x in 4:6=x:3

frozen [14]

Answer:

x = 2

Step-by-step explanation:

Express the ratio in fractional form

$\frac{4}{6}$ = $\frac{x}{3}$ ( cross- multiply )

6x = 12 ( divide both sides by 6 )

x = 2

4 0

3 years ago

If a is equal to 2 + root 3 upon 2 minus root 3 and b is equal to 2 minus root 3 upon 2 + root 3 then what is the value of a squ

natulia [17]

195

Step-by-step explanation:

hxjhxjxhcbhshshsgshnxgshjsjkabshzikndbxhdhjsnsvdvgxhxjsjjsv

8 0

2 years ago

A drawing is 12 inches wide and 10 inches long. Joe wants to enlarge each side by a scale factor of 0.3. What will be the area o

Andreyy89

Answer:

Area of the dilation = 10.8 square inches

Step-by-step explanation:

Given the original dimension

Length = 12inches

Width = 10inches

If they are dilated by a factor of 0.3

New length = 0.3 * 12 = 3.6in

New width = 0.3 * 10 = 3in

Area of the dilation = 3.6 * 3

Area of the dilation = 10.8 square inches

3 0

2 years ago