He initially had 300 boxes of candy. Every minute he sells 4 boxes. <br><br> WHAT IS THE SLOPE?

Alina and Asim share Rs 210 in the ratio 2:5. How much money did each of them get?

slavikrds [6]

Answer:

Step-by-step explanation:

Alina=2x---eq(1)

Asim=5x----eq---(2)

So,we get

2x+5x=210

7x=210

x=30

Now,Putting '30' in eq 1 and eq 2

2*30=60

5*30=150

Therefore Alina got 60 Rs while Asim got 150 Rs

5 0

3 years ago

Maria earns $80 per day working at a gas station. Write an algebraic expression to represent the amount of money she will earn i

Simora [160]

Answer:

The algebraic expression is 80d

Step-by-step explanation:

Maria's earnings per day = $80

Number of days= d

Total earnings for d days:

80d

Where,

80 is earnings per day

d is number of days

The algebraic expression is 80d

Algebraic equation is

y= 80d

Where

y= total income earned in d days

The expression is different from am equation because of the equal to sign(=)

3 0

3 years ago

Circle Z is intersected by AC←→ and CE←→.<br><br> What is the measure of AE?

Serggg [28]

36 = 1/2 (AE - 26)
72 = AE - 26
98 = AE

6 0

3 years ago

Read 2 more answers

Use the balanced scale to find the conversion factor that can be used to convert the number of blocks to the weight of the block

77julia77 [94]

Answer:

x = 2 1/4 or 2.25

Step-by-step explanation:

to start we can make an equation

9lbs = 4x

divide by 4

9/4 = x

x = 2.25 or 2 1/4

4 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

He initially had 300 boxes of candy. Every minute he sells 4 boxes. WHAT IS THE SLOPE?