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3 years ago

Can someone help with 2 and 3

Mathematics

2 answers:

hammer [34]3 years ago

7 0

For number 2 it's would be 12.8

Sorry I don't have time for number 3

RideAnS [48]3 years ago

6 0

I hope this helps you

1)

24/14=2.12/2.7=12/7

2)

4/8=4.1/2.4=1/2

3)

440/2=220.2/2.1=220/1

4)

x/6=6/24

x/6=6.1/4.6

x/6=1/4

x=6/4

x=3/2

x=1.5

5)

4/y+3=3/y-4

4 (y-4)=3 (y+3)

4y-16=3y+9

y=25

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A board is 12 feet long. If the board is divided into 12 equal sections, how long is each section?

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3 years ago

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

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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

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

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3 years ago

Find the missing section. Answers must be in fraction form.

frozen [14]

Answer:

$\frac{1}{10}$

Step-by-step explanation:

We have a circle that is split in three sections, two of which we know and we are asked to find the third missing section.

For the circle, we know that 4/5 and 1/10 is fit. Now we need the last one, to solve, we need to get the same denominator and see how much is missing.

Since 1/10 is our highest denominator, let's change 4/5 to have 10 as a denominator. Which would be through multiplying 5 to get 10.

What times 5 equals 10?

Now multiply both numerator and denominator by 2 to get our portion.

$\frac{4*2}{5*2}$

$\frac{8}{10}$

Now we have the same denominator, let's add our two fractions and see how much we have left.

8/10 + 1/10

9/10

We have 1/10 missing, therefore 1/10 is the answer.

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2 years ago