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2 years ago

Solve; 10ac-x/11=-3, for a

1 answer:

5 0

$\dfrac{10ac-x}{11}=-3\ \ \ \ |multiply\ both\ sides\ by\ 11\\\\10ac-x=-33\ \ \ |add\ x\ to\ both\ sides\\\\10ac=x-33\ \ \ |divide\ both\ sides\ by\ 10c\\\\\boxed{a=\frac{x-33}{10c}}$

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7+2n=8n-5 <br>n=? <br>whatsl does n equal

Westkost [7]

8n -2n=7+5

6n =12

n =2

this is the answer

8 0

2 years ago

Read 2 more answers

Please help me figure this out

Gnom [1K]

Answer:

True

(1) You can prove two triangles are similar using AA similarity.

(2)If side length of similar figure have ration $\frac{m}{n}$ then the ratio of surface areas will be: $\frac{m^2}{n^2}$

(3)If side length of similar figure have ratio of $\frac{2}{3}$ then perimeter will have ratio: $\frac{2}{3}$ .

False:

(1)If side length of similar figure have ration $\frac{m}{n}$ then the ratio of Volumes will be: $\frac{m^3}{n^3}$

(2)If side length of similar figure have ration $\frac{m}{n}$ then the ratio of surface area will be: $2\frac{m}{n}$

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2 years ago

find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2

Alexxandr [17]

The surface is parameterized by

$\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k$

and the normal to the surface is given by the cross product of the partial derivatives of $\vec s$ :

$\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}$

It looks like you're given

$\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}$

Then the normal vector is

$\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k$

Now, the point (2, 2, 0) corresponds to u and v such that

$\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}$

and solving gives $u = v = 1$ , so the normal vector at the point we care about is

$\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k$

Then the equation of the tangent plane is

$\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath + (z-0)\,\vec k\right) = 0$

$-4(x-2) + 2(y-2) - 4z = 0$

$\boxed{2x - y + 2z = 2}$

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1 year ago

In the figure shown, which part of angles must be complementary?

gladu [14]

Complementary angles are any 2 adjacent angles that make up 90°

knowing this, we see <COB is 90°, so now we just look for the 2 angles inside of it

<COD & <DOB

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2 years ago

Which data set has the smallest mean

kvv77 [185]

Answer:

Step-by-step explanation:

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2 years ago