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2 years ago

4x + 9y = 4

Mathematics

1 answer:

saw5 [17]2 years ago

7 0

Answer:

x=101/65, y=-16/65. (101/65, -16/65).

Step-by-step explanation:

4x+9y=4

9x+4y=13

----------------

9(4x+9y)=9(4)

-4(9x+4y)=-4(13)

------------------------

36x+81y=36

-36x-16y=-52

----------------------

65y=-16

y=-16/65

4x+9(-16/65)=4

4x-144/65=4

4x=4+144/65

4x=404/65

x=(404/65)/4=101/65

x=101/65, y=-16/65.

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What is the value of x in this equation

eduard

The the first one the answer should be -4

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2 years ago

Read 2 more answers

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

URGENT...

MatroZZZ [7]

The center of the dilation is the origin, and the scale factor is 2.

To find the coordinates of the points under this transformation, it is enough to multiply both coordinates of the points X, Y, and Z by 2.

Take a look at the dilation of point Z. To get to Z', both the x and y coordinate of Z must be multiplied by 2.

So, the coordinates are:

X'(8, 0); Y'(6, 4); Z'(4, -4).