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3 years ago

Does this set of ordered pairs represent a function? Why or why not?

Mathematics

1 answer:

Ivan3 years ago

8 0

Answer:

Step-by-step explanation:

Your line goes from 6 to -1 by x left and then from -1 to 3. Its impossible as function cannot have more than 1 y value per x value. So lools like its b. Hope this helps) Have a nice day)

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Please help! algebra 1 math! whoever gets right is Brainliest! (10 points)

Inga [223]

I believe that 7 5/4 is the answer.

7 0

3 years ago

Read 2 more answers

Find the sum of the first 6 terms of the following series, to the nearest integer.

natka813 [3]

Answer:

1008

Step-by-step explanation:

geometric sequence:

a=16

r=32/16=2

Sum of GS:

s=a(1-r^n)/(1-r)

S(6)=16(1-2^6)/(1-2)

=1008

8 0

3 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

What is the converse of r->(~q v p)

alexira [117]

Answer:

$r => (\neg q \ \lor p) \ \equiv \ q \ \land \ \neg p => \neg r$

Step-by-step explanation:

In general, remember that the converse of $a => b$ is $\neg b => \neg a$ , therefore in this case $a = r \ \ , b = \neg q \ \lor p$

so $r => (\neg q \ \lor p) \ \equiv \ q \ \land \ \neg p => \neg r$

8 0

3 years ago

The question is in the picture.

dangina [55]

Let's denote no of seats in first row with r1 , second row with r2.....and so on.

r1=5

Since next row will have 10 additional row each time when we move to next row,

So,

r2=5+10=15

r3=15+10=25

<u>Using the terms r1,r2 and r3 , we can find explicit formula</u>

r1=5=5+0=5+0×10=5+(1-1)×10

r2=15=5+10=5+(2-1)×10

r3=25=5+20=5+(3-1)×10

<u>So for nth row,</u>

rn=5+(n-1)×10

Since 5=r1 and 10=common difference (d)

rn=r1+(n-1)d

Since 'a' is a convention term for 1st term,

which is an explicit formula to find no of seats in any given row.

Using above explicit formula, we can calculate no of seats in 7th row,

r7=5+(7-1)×10

r7=5+(7-1)×10 =5+6×10

r7=5+(7-1)×10 =5+6×10 =65

which is the no of seats in 7th row.

8 0

3 years ago