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Ratling [72]
3 years ago
5

What is 2/x times 4z/9

Mathematics
1 answer:
Bond [772]3 years ago
4 0
Use photo math,, to save your points!
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Kaylee invested $910 in an account paying an interest rate of 2.6% compounded
Nikitich [7]

Answer:

$1,179

Step-by-step explanation:

Lets use the compound interest formula provided to solve this:

A=P(1+\frac{r}{n} )^{nt}

<em>P = initial balance</em>

<em>r = interest rate (decimal)</em>

<em>n = number of times compounded annually</em>

<em>t = time</em>

<em />

First, lets change 2.6% into a decimal:

2.6% -> \frac{2.6}{100} -> 0.026

Since the interest is compounded quarterly, we will use 4 for n. Lets plug in the values now:

A=910(1+\frac{0.026}{4})^{4(10)}

A=1,179.21

The account balance after 10 years will be $1,179

8 0
3 years ago
The model shows the addition expression 18 + 12.
Rashid [163]

Answer:

3(6+4) ._. hope this helps

3 0
3 years ago
Solve the equations. 2.5x−3.67=1.52
aleksklad [387]
Our current equation is:
2.5x -3.67 = 1.52.
To solve for x, we need to get x by itself and then simplify what we can from there.
2.5x -3.67 = 1.52
Add 3.67 to both sides to get 2.5x by itself.
2.5x = 5.19 is now our new equation.
Divide both sides by 2.5 to get x.
5.19/2.5 = 2.07
x = 2.07
I hope this helps!
5 0
3 years ago
What are the coordinates of the point where the total costs for both companies are the same? Question 1 options: (2, 60) (0, 0)
Molodets [167]

Answer:

0,0

Step-by-step explanation:

5 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
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