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3 years ago

Which is the best approximation for sqrt 55 A) 7 B) 7 2/5 C) 7 7/15 D) 8

Mathematics

1 answer:

Kitty [74]3 years ago

5 0

The best approximation for sqrt 55 would be: B) 7 2/5

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Select the correct answer.

nordsb [41]

Answer:

A. The graph is a line that goes through the points (9,0) and (0,6).

Step-by-step explanation:

Given

$4x + 6y = 36$

Required

Select the true options

$(a)\ (9,0)\ and\ (0,6)$

This implies that"

$(x_1,y_1)= (9,0)$

$(x_2,y_2)= (0,6)$

So, we have: $(x_1,y_1)= (9,0)$

$4x + 6y = 36$

$4 * 9 + 6 * 0 = 36 + 0 = 36$

So, we have: $(x_2,y_2)= (0,6)$

$4 * 0 + 6 * 6 = 0 + 36 = 36$

Hence (a) is true.

3 0

3 years ago

What is the volume? h 12 cm Round your answer to the nearest tenths:

denis23 [38]

Answer: So if its a sphere the answer its 7238.2.

If its a cylinder the answer is 371.7.

If its a triangular shape the answer is

123.9.

6 0

3 years ago

Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v

k0ka [10]

I'll use subscript notation for brevity, i.e. $\frac{\partial f}{\partial x}=f_x$ .

By the chain rule,

$z_u=z_xx_u+z_yy_u$

$z_v=z_xx_v+z_yy_v$

$z_w=z_xx_w+z_yy_w$

We have

$z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}$

and

$\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}$

When $u=1,v=1,w=0$ , we have

$\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}$

and the partial derivatives take on values of

$\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}$

So we end up with

$\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}$

3 0

3 years ago

Point W is on line segment VX. given VW=3 and VX=14, determine the length WX

r-ruslan [8.4K]

Answer:11

Step-by-step explanation:14-3=11

4 0

4 years ago

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Elina [12.6K]

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3 years ago