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3 years ago

Name a fourth point in plane TUY. x Y W Z

Mathematics

1 answer:

Elan Coil [88]3 years ago

6 0

Answer:

Step-by-step explanation:

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If f(x) = 3x - 9 and g(x) = x², what is (gºf)(5)?

Xelga [282]

Answer:

Step-by-step explanation:

Here, f(x) is the input to g(x), forming a composite function.

To evaluate (gºf)(5),

1) find the value of f(5). It is f(5) = 6.

2) Use this result as the input to g(x): g(6) = (6)^2 = 36

3 0

3 years ago

What is the y-intercept, axis of symmetry and vertex of the following function.

tamaranim1 [39]

To find the y intercept you must substitute 0 into x.

F(x) = (-3x0) - (6x0) - 5
F(x) = -5

To find the axis of symmetry and vertex we use the formula -b/2a to find the x coordinate

x = 6/(-3x2)
x = 6/-6
x = -1

Now we need to find the y coordinate so we sub -1 into x.

F(x) = (-3x-1) - (6x-1) - 5
F(x) = 3 + 6 - 5
F(x) = 4

The axis of symmetry and vertex are (-1,4) and the y coordinate is -5. Hope this helps.

5 0

2 years ago

Read 2 more answers

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

Find the Quetion of the line that passes through the points (1,6) and (5,2)

NemiM [27]

Answer:

$y=-x+7$

Step-by-step explanation:

Equation of straight line is given as

$y-y_1=m(x-x_1)$

where slope m is calculated as

$m=\frac{y_2-y_1}{x_2-x_1}$

Here 

$(x_1, y_1)=(1, 6) \ \ and \ \ (x_21, y_2) = (5, 2)$

$m = \frac{2 - 6}{5 - 1}\\m=\frac{- 4}{4} \\m=-1$

Equation of line that passes through the points (1,6) and (5,2)

$y-6=-1(x-1)\\y=-x+7$

5 0

3 years ago

Even has a bag of 14 jelly beans. Four of his jelly beans are black, five are red, and five are purple. What is the ratio of the

nirvana33 [79]

Answer: 14:4

Step-by-step explanation:

The amount of jelly beans: 14

The amount of black jelly beans: 4

all beans: black beans

14:4

6 0

2 years ago