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Ulleksa [173]
3 years ago
13

In a scale drawing of a rectangular deck, the

Mathematics
1 answer:
KonstantinChe [14]3 years ago
6 0

Answer:

45

Step-by-step explanation:

10/4x3=7.5

8/4x3=6

6x7.5=45yards

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Jason is playing a trivia game with his friends. At the end of each round, his score updates to the square of 1 less than the pr
mezya [45]

Answer:

8+n^2

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Match each operation involving f(x) and g(x) to its answer
eimsori [14]

Answer:

(g-f) (-1)= sqrt(15)

(f/g)(-1)= 0

(g+f)(2)=sqrt(3)-3

(g*f)(2)=-3*sqrt(3)

Step-by-step explanation:

We have to eval the expressions given in the point indicated.

Lets start by the first equation

(g-f)(-1)= g(-1) - f(-1)= \sqrt{11-4*(-1)}    - 1 +(-1)^{2} = \sqrt{15}

Now, lest continue with the others

(f/g)(-1)= f(-1)/g(-1)= (1-1)/sqrt(15)=0

(g+f)(2)=g(2)+f(2)=sqrt(3)-3

(g*f)(2)=g(2)*f(2)=sqrt(3)*(-3)=-3sqrt(3)

3 0
3 years ago
Read 2 more answers
Find the midpoint of NP⎯⎯⎯⎯⎯ given N(2a, 2b) and P(2a, 0).
geniusboy [140]

Answer:

(2a, b )

Step-by-step explanation:

Given the endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

[ \frac{1}{2}(x₁ + x₂ ), \frac{1}{2}(y₁ + y₂ ) ]

Here (x₁, y₁ ) = N(2a, 2b) and (x₂, y₂ ) = P(2a, 0), thus

midpoint = [ \frac{1}{2}(2a + 2a), \frac{1}{2}(2b + 0 ) ] = (2a, b )

4 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
Can Somebody Solve This Inequality? Show Work
Kaylis [27]

0

6 0
4 years ago
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