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2 years ago

Every 2/3 hour, Harris can sew 1/6 pair of jeans.

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

6 0

.25 or 1/4
i hope that this answer helps you today

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5. One canned juice drink is 15% orange juice; another is 10% orange juice. How many liters of each should be mixed together in

hram777 [196]

X: liters of 15% orange juice
y: liters of 10% orange juice

x+y=5L

y=5L - x

x liters of 15% orange juice plus y liters of 10% orange juice makes 5 liters of 12% orange juice in equation form:

0.15x+0.10y = 0.12(5L)

0.15x+0.10(5L-x)=0.12(5L)

0.15x+0.5-0.10x=0.6

0.05x=0.1L

x=2L

so y = 3L

so you need 2 liters of 15% orange juice and 3 liters of 10% orange juice

4 0

2 years ago

Can some one help me plsss!!?

Dahasolnce [82]

Ha uuuu un día y que ha dado la oportunidad para el teléfono

3 0

1 year ago

Hey can y’all help me with this one

neonofarm [45]

Answer:

um what I do is I go from like if I was doing 12 I would go 1 times 12 and then 2 times 6 and 3 times 4 and then I would stop cause it's repetitive if that makes sense

8 0

2 years ago

Rewrite each expression in the form kkxxnn, where kk is a real number and nn is an integer. Assume x ≠ 0.

labwork [276]

Answer:

The expression will be $2x^5\times x^{10}=2x^{5+10}=2x^{15}$

Step-by-step explanation:

We have given the expression $2x^5\times x^{10}$

We have to write this expression in kkxxnn form , here kk is real number and nn is integer

We know the property of exponent that when two function are multiplied thn their exponent are added

So $2x^5\times x^{10}=2x^{5+10}=2x^{15}$

Here kk = 2 and nn = 15

7 0

2 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