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2 years ago

Write an equation that represents the graph of the line shown in the coordinate plane.

Mathematics

1 answer:

natulia [17]2 years ago

4 0

Answer:

y = 2x.

Step-by-step explanation:

The slope of the line = (y2-y1)(x2-x1)

= (8 - (-4)) / (4 - (-2))

= 12/6

= 2.

Now the y intercept is at (0, 0) so

the equation is y = 2x + 0

Answer: y = 2x.

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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

2 years ago

HELP ASAP PLS PLS PLS !

Maslowich

Answer:

(b) the unknown angle is 80°

Step-by-step explanation:

The unkown angle is:

[2(x-10)]°

Substituting "x" by 50 (part a)

[2(50-10)]°=[2(40)]°=80°

6 0

3 years ago

sam is 27 years old. his age is 6 years greater than 3 times Brandon's age. which method can be used to determine Brandon's age?

dezoksy [38]

Answer:

Step-by-step explanation:

6 times 3 = 18

27- 18 = 9 brandon is 9

3 0

3 years ago

Read 2 more answers

How to solve this question

VMariaS [17]

<span>The variance method is as follows.

-Sum the squares of the values in data set, and then divide by the number of values in data set
- From that, subtract the square of the mean (add all values and divide by number of values in the data set)

Our variance is

<span> $\displaystyle\sigma^2 = \frac{2^2 + 5^2 + m^2}{3} - \left(\frac{2 + 5 + m}{3}\right)^2$

Since variance has to be 14, we set $\sigma^2 = 14$ and solve for m

$14= \frac{4 + 25 + m^2}{3} - \left(\frac{7 + m}{3}\right)^2\ \Rightarrow \\ \\
14 = \frac{29}{3} + \frac{1}{3}m^2 - \frac{1}{9}(7+m)^2 \\ \\
14 = \frac{29}{3}+ \frac{1}{3}m^2 - \frac{1}{9}(49 + 14m + m^2) \\ \\
14 = \frac{29}{3}+ \frac{1}{3}m^2 - \frac{49}{9}- \frac{14}{9}m- \frac{1}{9}m^2 \\ \\
0 = \frac{-88}{9} -\frac{14}{9}m + \frac{2}{9}m^2
$

quadratic formula

$m = \displaystyle\frac{-b \pm \sqrt{b^2 -4ac}}{2a} \\
m = \frac{-(-\frac{14}{9}) \pm \sqrt{\left(-\frac{14}{9}\right)^2 - 4(2/9)(-88/9)} }{2(2/9)} \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{196}{81} + \frac{704}{81} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{900}{81} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{100}{9} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \frac{10}{3} }{\frac{4}{9} } \\
m = 11, -4$

-4 doesnt' work as it is not a positive integer

m = 11

</span></span>

5 0

2 years ago

The clock below shows time that lance got home from school. 3:32. What time did he get home A: seventeen minutes before three B:

navik [9.2K]

Answer:

If Lance got home from school at 3:32 p.m. A and B both represent:

A: seventeen minutes before 3 would be before 3 o'clock even happens

B: 28 minutes before 4 would make more sense, since 3:32 is before 4 o'clock

Lance got home 28 minutes before 4

Hope this helps ;)

5 0

2 years ago