All categories

3 years ago

Quadrilateral ABCD has the following coordinates:

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

4 0

Answer:

A¹ (3 , -1) , B¹(0,-8) , C¹(8,-9) , D¹(7,-2)

Step-by-step explanation:

Explanation:-

Given coordinates are

A(1,3) ,B(8,0) ,C( 9,8) ,D(2,7)

The coordinates are rotating the shape 90° clockwise then the transformed to (x,y)→ (y ,-x)

The new coordinates

A(1,3) → A¹ (3 , -1)

B(8,0) → B¹(0,-8)

C(9,8) → C¹(8,-9)

D(2,7) → D¹(7,-2)

You might be interested in

Please answer this correctly

GREYUIT [131]

Answer:

$8.82

Step-by-step explanation:

4.6 (1.15) + 0.5 (1.12) + 2.25 (1.32)

5.29 + 0.56 + 2.97

8.82

8 0

3 years ago

What is equivalent to (x+8)(x+4)? 1.x^2+12x32 2.2x^2+10x+12 3.2x^2+12x+16

Elza [17]

Answer:

1. x^2+12x+32

Step-by-step explanation:

(x+8)(x+4)

x(x+4)+8(x+4)

x^2+4x+8x+32

x^2+12x+32

3 0

2 years ago

Read 2 more answers

In which quadrants do solutions for the inequality y is less than or equal to two sevenths times x plus 1 exist?

hram777 [196]

Answer:

y will be in every single quadrant

Step-by-step explanation:

So we have the equation $y\leq \frac{2}{7} x+1$ first we will have to look at the equation. It says that y is less than or equal to $\frac{2}{7} x+1$ since y is less than $\frac{2}{7} x+1$ the only place the shaded area where y can be is under the line that is drawn be the equation. When the equation is graphed the y-intercept will be on positive 1 it since slope is rise over run it will look something like the file attached to this. so under the line you can see every single quadrant so that is why it would be that way

5 0

3 years ago

HELP PLS THIS IS HARD WLH

AnnyKZ [126]

Answer:

Step-by-step explanation:

A'(-2*3/2=-3,3*3/2=4.5), B'(-6,-6), C'(6,0), D'(7.5,6)

8 0

2 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago