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cluponka [151]
4 years ago
9

Jed's has $45 for the amusement park. He spent $17 for admission into the park. How much money does he have left after he paid f

or his admission?
Mathematics
2 answers:
Alik [6]4 years ago
5 0

45-17=28

hope this helps

seropon [69]4 years ago
5 0
To find out the answer all you have to do is 45-17=28
So he would have $28 left
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Suppose that the water level of a river is 34 feet and that it is receding at a rate of
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In 2009, a tree was 10 feet tall. In 2012, the tree was 16 feet tall. Construct a linear model that represents the height, h, of
HACTEHA [7]

Answer:

h = 2t + 4

Step-by-step explanation:

Solution:-

- The height of the tree = h

- The number of years elapsed after 2006 = t

- height h1 = 10 ft, t1 = ( 2009 - 2006 ) = 3 yrs

- height h2 = 16 ft, t2 = ( 2012 - 2006 ) = 6 years

- The linear model of tree height "h" as a function of time "t" years after 2006 can be expressed in the form:

                           h = m*t + c

Where,               m = The rate of change of height

                          c = The initial height of tree in 2006.

- We will use the given data to evaluate constant "m".

                         Rate =  m = ( h2 - h1 ) / ( t2 - t1)

                         m = ( 16 - 10 ) / ( 6 - 3 )

                         m = 6 / 3 = 2

-  Then use "m" and given set of data to evaluate the initial height of tree "c" in year 2006.

                          h = 2*t + c

                          h1 = 2*t1 + c

                          c = 10 - 2*3 = 4 ft

- The linear model for the height of tree "h" as fucntion of time "t" years after 2006 will be:

                          h = 2t + 4

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Answer is B.

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(a) The plane y + z = 13 intersects the cylinder x2 + y2 = 25 in an ellipse. Find parametric equations for the tangent line to t
klemol [59]

Answer:

Step-by-step explanation:

We have a curve (an ellipse) written as the system of equations

\begin{cases} y+z &= 13\\ x^2+y^2 &= 25\end{cases}.

And we want to calculate the tangent at the point (3,4,9).

The idea in this problem is to consider two variables as functions of the third. Usually we consider y and z as functions of x. Recall that a curve in the space can be written in parametric form in terms of only one variable. In this case we are considering the ‘‘natural’’ parametrization (x, y(x), z(x)).

Recall that the parametric equation of a line has the form

r(t)=\begin{cases} x(t) &= x_0 + v_1t \\ y(t) &= y_0 +v_2t\\ z(t) &= z_0 +v_3t \end{cases},

where (x_0,y_0,z_0) is a point on the line (in this particular case is (3,4,9)) and (v_1,v_2,v_3) is the direction vector of the line. In this case, the direction vector of the line is the tangent vector of the ellipse at the point (3,4,9).

Now, if we have the parametric equation of a curve (x, y(x), z(x)) its tangent line will have direction vector (1, y'(x), z'(x)). So, as we need to calculate the equation of the tangent line at the point (3,4,9) = (3, y(3), z(3)), we must obtain the tangent vector (1, y'(3), z'(3)). This part can be done taking implicit derivatives in the systems that defines the ellipse.

So, let us write the system as

\begin{cases} y(x)+z(x) &= 13\\ x^2+y^2(x) &= 25\end{cases}.

Then, taking implicit derivatives:

\begin{cases} y'(x)+z'(x) &= 0 \\ 2x+2y(x)y'(x) &= 0\end{cases}.

Now we substitute the values x=3 and y(3)=4, and we get the system of linear equations

\begin{cases} y'(3)+z'(3) &= 0 \\ 2\cdot 3+2\cdot 4y'(x) &= 0\end{cases},

where the unknowns are y'(3) and z'(3).

The system is

\begin{cases} y'(3)+z'(3) &= 0 \\ 6+8y'(x) &= 0\end{cases},

and its solutions are

y'(3) = -\frac{3}{4} and z'(3) = \frac{3}{4}.

Then, the direction vector of the tangent is

(1, -\frac{3}{4}, -\frac{3}{4}).

Finally, the tangent line has parametric equation

r(t)=\begin{cases} x(t) &= 3 + t \\ y(t) &= 4 -\frac{3}{4}t\\ z(t) &= 9 +\frac{3}{4}t \end{cases}

where t\in\mathbb{R}.

7 0
4 years ago
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