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2 years ago

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a parrallelogram is graphed on a coordinate plane so the two points are in the first quadrant and two points are in the third qu

adrant. What are the possible coordinates of the vertices of the parallelogram?

1 answer:

Rasek [7]2 years ago

8 0

A quadrilateral is said to be a parallelogram

1.Opposite sides are equal and parallel.

2. Diagonal bisect each other.

3. Opposite angles are equal.

It is given that , a Parrallelogram is graphed on a coordinate plane so the two points are in the first quadrant and two points are in the third quadrant.

Suppose ABCD is a Parallelogram.Then AB=CD and AD=BC.

Given, Vertices A,B lies in first Quadrant and Vertices C and D lies in Third Quadrant.Then Vertices of Parallelogram ABCD are

A=( x, y) and B=( y, x)

Then, C= (- x,- y) and D= (- y,- x), Arranged in Alphabetical order, that is A,B and C and D.

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Find the derivative of the function at P 0 in the direction of A. f(x,y,z) = 3 e^x cos(yz), P0 (0, 0, 0), A = - i + 2 j + 3k

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The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

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Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$

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Step-by-step explanation:

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Solve for a and b<br> (6a-3)=(7b-10)

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Answer:

for A= $\frac{7b-7}{6}$ and for B= $\frac{6a+7}{7}$

Step-by-step explanation:

For A

$\left(6a-3\right)=\left(7b-10\right)$

$6a-3+3=7b-10+3$

$6a=7b-7$

$\frac{6a}{6}=\frac{7b}{6}-\frac{7}{6}$

$a=\frac{7b-7}{6}$

For B

$\left(6a-3\right)=\left(7b-10\right)$

$7b-10=\left(6a-3\right)$

$7b-10=6a-3$

$7b-10+10=6a-3+10$

$7b=6a+7$

$\frac{7b}{7}=\frac{6a}{7}+\frac{7}{7}$

$b=\frac{6a+7}{7}$

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