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2 years ago

Expand the following expression

Mathematics

1 answer:

masya89 [10]2 years ago

5 0

Answer:

$\frac{5}{4} \times 4x + \frac{5}{4} \times \frac{3}{4} = 5x + \frac{15}{16}$

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HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

saul85 [17]

Answer:

Option B is the correct answer

Step-by-step explanation:

A line segment is a line drawn with two end points.

1.) Option A is a point

2.) Option B is a line segment

3.) Option C is a ray

4.) Option D is a line

Hope this helps!

8 0

2 years ago

What is the answer?

lisov135 [29]

f is the answer hope it helps

4 0

3 years ago

3-10-(-15) Show and explain work

Genrish500 [490]

Answer:

3-10-(15) equals 8.

Step-by-step explanation:

3-10 equals -7. Then -(-15) becomes a positive because two negatives make a positives producing +15 so it becomes -7+15 which finally becomes 8.

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

Suppose Carol Danvers invested $1,000 into an account paying 6% annual interest compounded

Jet001 [13]

Answer:

$ 1,060.00

Step-by-step explanation:

A = $ 1,060.00

A = P + I where

P (principal) = $ 1,000.00

I (interest) = $ 60.00

Compound Interest Equation

A = P(1 + r/n)^nt

Where:

A = Accrued Amount (principal + interest)

P = Principal Amount

I = Interest Amount

R = Annual Nominal Interest Rate in percent

r = Annual Nominal Interest Rate as a decimal

r = R/100

t = Time Involved in years, 0.5 years is calculated as 6 months, etc.

n = number of compounding periods per unit t; at the END of each period

6 0

3 years ago