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3 years ago

5

Given triangle ABC with coordinates A(−2, 7), B(−2, 4), and C(−4, 3), and its image A′B′C′ with A′(2, 3), B′(−1, 3), and C′(−2,

1), find the line of reflection.
The line of reflection is at y =

1 answer:

Setler [38]3 years ago

6 0

Answer:

y = x + 5

Step-by-step explanation:

given an obtuse scalene triangle (∆ABC) with vertices A(-2,7), B(-2,4), and C(-4,3) (representing the preimage of the reflection). And congruent triangle (∆A'B'C') with vertices A'(2,3), B'(-1,3), and C'(-2,1) (representing the image of the reflection / triangle after the reflection).

The easiest way to look at this is by taking the inverse of each point on the preimage (swapping y and x coordinates) and the adding 5 to the y and subtracting 5 from the x to get the image.

You can test this, and you will see that this will match with vertices of the image.

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Paul has a number of counters two less than seven

valentinak56 [21]

The answer is 5 cause two less then 7 is 5.

4 0

2 years ago

Nine less than a number is no more than 8 and is no less than 3.

snow_lady [41]

3 is less than or equal to x-9 is less than or equal to.

3<= x-9<=8

<= these mean less than or equal to.
hope I helped you!

3 0

3 years ago

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

3 years ago

There were 512 cups of water in a cooler before a soccer game. After the soccer game there were 384 cups of water in the cooler.

shusha [124]

Answer:

C

Step-by-step explanation:

you need to find how many cups of water were used in the game:

512-384=128

and then you can begin to calculate the number after that:

128/16=8(gallons)

hope it'll help! have a good day!

5 0

2 years ago

Members of video game platform X-

Natalija [7]

Answer:

When number of games downloaded is at least 3

Step-by-step explanation:

Let x = number of games downloaded

Members = 50 + 2x

Non members = 20x

50 + 2x = 20x

50 = 20x - 2x

50 = 18x

x = 50/18

x = 2.78

At x = 2.78, both members and non members are paying the same amount

For members to make profits, the value of x must be greater than 2.78

Approximately x = 3

Check:

Members = 50 + 2x

= 50 + 2(3)

= 50 + 6

= 56

Non members = 20x

= 20(3)

= 60

3 0

2 years ago