All categories

3 years ago

How can one fourthx − 3 = one halfx + 8 be set up as a system of equations?

Mathematics

1 answer:

Delvig [45]3 years ago

6 0

<span>The answer is B.

The reason being is that in this option they have set both sides of the equation separately equal to y. Since they are equal to each other, they would both have to be equal to an unknown y. This would look like this.
y = 1/4x - 3
y = 1/2x + 8.

Then to remove the fractions in each, they multiplied by the denominator associated with x. So, you multiply the first equation by 4.
4y = x - 12.
Then they solve for the numerical value at the end.
4y - x = -12.

You can do the same with the second equation.
y = 1/2x + 8
2y = x + 16
2y - x = 16

These are now the two equations above. </span>

You might be interested in

They or both correct

yarga [219]

Answer:

wut.

Step-by-step explanation:

7 0

3 years ago

a patient is receiving 1 liter of IV fluids every 8 hours at that rate how much will he recieve in 3 hrs

ASHA 777 [7]

Answer:

Step-by-step explanation:

1litres=8

x=3

x=0.375 litres

3 0

3 years ago

I need the answer Algebra question

Sav [38]

He had 32 dollars at the beginning of the day

5 0

3 years ago

How do you find the derivative of 4

Nutka1998 [239]

by rule the derivative of a constant is 0

so the answer for this is 0

5 0

3 years ago

Compute the differential of surface area for the surface S described by the given parametrization.

AysviL [449]

With $S$ parameterized by

$\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle$

the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

We have

$\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle$

$\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle$

with cross product

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle$

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}$

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=e^u\sqrt{u^2+v^2+e^{2u}}$

So we have

$\mathrm dS=\boxed{e^u\sqrt{u^2+v^2+e^{2u}}\,\mathrm du\,\mathrm dv}$

8 0

3 years ago