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3 years ago

9

CAN SOMEONE PLEASE HELP ME!!!!

2 answers:

padilas [110]3 years ago

6 0

Answer:

The third answer aka C

Step-by-step explanation:

vitfil [10]3 years ago

6 0

Answer: The first option is the correct answer

Step-by-step explanation: It is not a function because it does not pass the vertical line test.

The vertical line test is a test that determines whether or not a relation is a function. To perform the vertical line test you must go through every point and look DOWN the line in which the point lies. If there is another point on the same Y axis then it does not pass the vertical line test which means its not a function.

HOPE THIS HELPS!

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Plz help with math!!!

Orlov [11]

5y+2 i think because those are parallel lines:)

5 0

3 years ago

What is the value of k if 7k" minutes past nine" is the same as 8k" minutes to ten"?

schepotkina [342]

Answer:

4 minutes

Step-by-step explanation:

The value of 7k "minutes past nine" is the same as 8k" minutes to ten".

Since there are 60 minutes in an hour and 9 and 10 fall in between the same hour, this means that:

7k = 60 - 8k

7k + 8k = 60

15k = 60

k = 60 / 15

k = 4 minutes

5 0

3 years ago

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

Select all operations under which the polynomials 5x - 1 and 3x - 10 are not closed.

sammy [17]

It's division.
Hope I helped!

7 0

3 years ago

write the slope - intercept equation of the line that passes through the point (6, "-1)" and has a slope of "-1/3"

TEA [102]

Answer:

y = - $\frac{1}{3}$ x + 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = - $\frac{1}{3}$ , then

y = - $\frac{1}{3}$ x + c ← is the partial equation

To find c substitute (6, - 1 ) into the partial equation

- 1 = - 2 + c ⇒ c = - 1 + 2 = 1

y = - $\frac{1}{3}$ x + 1 ← equation of line

8 0

2 years ago