Ask question

Ask question

All categories

3 years ago

7

A salesperson's weekly paycheck is 40% more than a second salesperson's paycheck. The two paychecks total $1075. Find the amount

of each paycheck.

1 answer:

BlackZzzverrR [31]3 years ago

3 0

X is gonna be the normal paycheck

0.45x+x=1075

= 0.40x+x=1075

= 1.4x=1075

x= 1075/1.4

x=767.85 the normal paycheck price

1075.00-767.85= 307.15

so a saleperson weekly paycheck is 307.15 dollars more

hope this helps

You might be interested in

Can someone help with this question?

Firlakuza [10]

Answer:

c=5.50b

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Help me asap..........:

Oduvanchick [21]

Answer:

J. P+7 less than or equal to 50

Step-by-step explanation:

You don't even need to look at the left side of the equation. There is a maximum of $50 you can spend so you can spend $50 or less than $50

3 0

3 years ago

Can you combine like terms without formally showing the distributive property? Explain.

blsea [12.9K]

Yes you can! Say you have 5(2 + x) + 3x. The answer would be 10+8x because when you use the distributive property you will end up getting 5*2 5*x and once you multiply, it's obvious 5 times 2 is 10 and 5 times x will equal 5x. Then you add 5x and 3x and get 8x. So, the simplified expression here is 10+8x. When you have this as an answer it's proving your answer right. You can without using the distributive property. We can have 10+5x+3x and it will be your simplified expression. It's the same thing with equations too.

6 0

3 years ago

Please help me with this question!!!!!

Naddika [18.5K]

Answer:

Equation of line in slope-intercept form that passes through (4, -8) and is perpendicular to the graph $y= \frac {2}{5} x - 3$ is below

$y = - \frac {5}{2} x + 2$

Step-by-step explanation:

Slope of the equation $y= \frac{2}{5} x -3$ is $m_1 = \frac{2}{5}$

Since slopes of perpendicular lines are negative reciprocal of each other, therefore slope of other line is given as

$m_2 = - \frac {1}{m_1} = - \frac {5}{2}$

Equation of line in point slope form is given as

$y-y_1=m_2(x-x_1)$

Here (x1, y1) = (4, -8)

$y+8 = - \frac{5}{2} (x-4)$

Simplifying it further

$y = - \frac {5}{2} x + \frac {5}{2} 4 - 8$

$y = - \frac {5}{2} x + 2$

6 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