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2 years ago

7

Pls help & also give an easy explanation thank youuuuu

1 answer:

Alex2 years ago

3 0

Given

A digital picture frame with a border of 3 cm. The actual length of the frame is x

Answer

a) The actual side of the picture is x-3

Area of picture

$=(side)^2=(x-3)^2$

b) Area of frame

$x^2$

c) Area of border = Area of frame - area of picture

$x^2-(x-3)^2$

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Andrew is riding his bike. He biked a distance of 14 miles at a rate of 7 miles per hour. Using the distance formula, d = it, so

Assoli18 [71]

Answer:

120 minutes

Step-by-step explanation:

D=it 14/7 is 2, multiply 2 by 60 you get 120

5 0

3 years ago

Which of the following is a factor of x2 − 21x + 110?

ludmilkaskok [199]

X² - 21x + 110
first you break it up into two parenthesis & since x is squared you need two x's
since 110 is positive and 21 is negitive, both of your signs need to be negitive
(x - ?)(x - ?)
then you need to think! what are the factors of 110?
Which of these factors add up to 21? those factors go in the parenthesis

Does this Help? Let me know if you need anything!

8 0

3 years ago

Read 2 more answers

Someone please help I want to double check PLEASEEEEE HELLPPPP

nikitadnepr [17]

Answer:

correct

Step-by-step explanation:

6 0

3 years ago

Which of the following is the correct factorization of the polynomial below?

ruslelena [56]

Answer:A

Step-by-step explanation:

7 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

4 years ago