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3 years ago

How do you solve for y?

Mathematics

1 answer:

Alchen [17]3 years ago

5 0

To find 'y', you must get it by itself in the equation. In order to do so, you need to get rid of other values on that side of the equation, while still making the expression equal the same. You need to "balance the equation". To do so, subtract 2 from both sides of the equation.

y + 2 = x
- 2 -2
y = x - 2

Your simplest answer for this problem would be y = x - 2

Hope this helps!

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What is the soulution set of the equation 3x^2=48

valentina_108 [34]

Answer:

x = ± 4

Step-by-step explanation:

given 3x² = 48 ( divide both sides by 3 )

x² = 16 ( take the square root of both sides )

x = ± $\sqrt{16}$ = ± 4

x ∈ {- 4, 4 }

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Nicholas has a salt shaker in the shape of a cylinder. It has a radius of 8 mm and a height of 30 mm. What is the lateral surfac

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last summer, at camp okey-fun-okey, the ratio of the number of boy campers to girl campers was 8:7. If there were a total of 195

Evgen [1.6K]

The ration of boy campers to total campers is 8:15, and the ratio of girl campers to total campers is 7:15. Using this information, we can answer this question by setting up proportions.

For boys:

$\frac{8}{15} =\frac{x}{195}$

*Cross multiply*

15x=1560

*Divide both sides by 15*

x=104

There are 169 boy campers.

For girls:

$\frac{7}{15} =\frac{x}{195}$

*Cross multiply*

15x=1365

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There are 91 girl campers.

Hope this helps!!

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Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−

aleksandrvk [35]

The tangent to $C$ through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces $S$ and $T$ at that point.

Let $f(x,y,z)=x^2+2y^3+3z^4$ . Then $S$ is the level curve $f(x,y,z)=6$ . Recall that the gradient vector is perpendicular to level curves; we have

$\nabla f(x,y,z)=(2x,6y,12z^2)$

so that the gradient of $f$ at (1, 1, 1) is

$\nabla f(1,1,1)=(2,6,12)$

For the surface $T$ , we have

$\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0$

so that $\vec r(1,0)=(1,1,1)$ . We can obtain a vector normal to $T$ by taking the cross product of the partial derivatives of $\vec r(u,v)$ , and evaluating that product for $u=1,v=0$ :

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)$

$\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)$

Now take the cross product of the two normal vectors to $S$ and $T$ :

$(2,6,12)\times(1,-1,2)=(24,8,-8)$

The direction of vector (24, 8, -8) is the direction of the tangent line to $C$ at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by $t\in\Bbb R$ . Then adding (1, 1, 1) shifts this line to the point of tangency on $C$ . So the tangent line has equation

$\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)$

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3 years ago