barb ate six cheeseburgers in the last four months. How many do you perdict she will eat in two years?

22 Points!! I NEED Help ASAP plz!!!!

Schach [20]

Answer:

The correct option is B.

Step-by-step explanation:

The vertices of preimage are L(2,-2), M(4,-2), N(1,-3) and O(5,-3).

The vertices of image are L'(-3,-2), M'(-3,-4), N'(-4,-1) and O'(-4,-5).

The relationship between preimage and image is defined as

$(x,y)\rightarrow (y-1,-x)$

In option A, 90-degree counterclockwise rotation about the origin followed by a translation 1 unit to the left.

If a figure rotated 90-degree counterclockwise, then

$(x,y)\rightarrow (-y,-x)$

Translation 1 unit to the left,

$(x,y)\rightarrow (-y-1,-x)$

Therefore option A is incorrect.

In option B, 270-degree counterclockwise rotation about the origin followed by a translation 1 unit to the left.

$(x,y)\rightarrow (y-1,-x)$

Therefore option B is correct.

In option C, a translation 1 unit to the left followed by a 270-degree counterclockwise rotation about the origin.

$(x,y)\rightarrow (y,-(x-1))$

Therefore option C is incorrect.

In option D, a translation 1 unit to the left followed by a 90-degree counterclockwise rotation about the origin.

$(x,y)\rightarrow (-y,x-1)$

Therefore option D is incorrect.

5 0

2 years ago

Read 2 more answers

What are soul toon to the equation below 4x^2-52x+169=122

Andrews [41]

hope this helps! let me know if i am wrong!

6 0

2 years ago

Find the general solution of the equation: (y^2-2xy)dx + (2xy-x^2)dy = 0

lys-0071 [83]

The ODE is exact, since

$\dfrac{\partial(y^2-2xy)}{\partial y}=2y-2x$

$\dfrac{\partial(2xy-x^2)}{\partial x}=2y-2x$

so there is solution $f(x,y)=C$ such that

$\dfrac{\partial f}{\partial x}=y^2-2xy$

$\dfrac{\partial f}{\partial y}=2xy-x^2$

Integrating both sides of the first PDE wrt $x$ gives

$f(x,y)=xy^2-x^2y+g(y)$

Differentiating both sides wrt $y$ gives

$\dfrac{\partial f}{\partial y}=2xy-x^2+\dfrac{\mathrm dg}{\mathrm dy}=2xy-x^2$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

Then the solution to the ODE is

$f(x,y)=\boxed{xy^2-x^2y=C}$

# # #

Alternatively, we can see that the ODE is homogeneous, since replacing $x\to tx$ and $y\to ty$ reduces to the same ODE:

$((ty)^2-2(tx)(ty))\,\mathrm d(tx)+(2(tx)(ty)-(tx)^2)\,\mathrm d(ty)=0$

$t^3(y^2-2xy)\,\mathrm dx+t^3(2xy-x^2)\,\mathrm dy=0$

$(y^2-2xy)\,\mathrm dx+(2xy-x^2)\,\mathrm dy=0$

This tells us we can solve by substituting $y(x)=xv(x)$ , so that $\mathrm dy(x)=x\,\mathrm dv(x)+v(x)\,\mathrm dx$ , and the ODE becomes

$(x^2v^2-2x^2v)\,\mathrm dx+(2x^2v-x^2)(x\,\mathrm dv+v\,\mathrm dx)=0$

$v(v-2)\,\mathrm dx+(2v-1)(x\,\mathrm dv+v\,\mathrm dx)=0$

$3v(v-1)\,\mathrm dx+x(2v-1)\,\mathrm dv=0$

which is separable as

$\dfrac{1-2v}{3v(v-1)}\,\mathrm dv=\dfrac{\mathrm dx}x$

Integrating both sides gives

$-\dfrac13\ln|v(1-v)|=\ln|x|+C$

$v(1-v)=\dfrac C{x^3}$

and solving in terms of $y(x)$ ,

$\dfrac yx\left(1-\dfrac yx\right)=\dfrac C{x^3}$

$xy(x-y)=C$

$\boxed{x^2y-xy^2=C}$

6 0

2 years ago

A cube with side lengths of 4 cm has a density of 3 grams/cubic centimeters. The mass of the cube is _____ grams?

Alex17521 [72]

9514 1404 393

Answer:

21 1/3 grams

Step-by-step explanation:

The mass is the product of the volume and the density. The volume of a cube is the cube of its edge dimension.

M = Vρ

M = (4 cm)³×(3 g/cm³) = 64/3 g

The mass of the cube is 64/3 = 21 1/3 grams.

8 0

2 years ago

Find the slope of the line

Rama09 [41]

the slope is the number that x is multiplied by so the answer is C)7

8 0

2 years ago