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uranmaximum [27]
3 years ago
6

You are going on vacation. You have $105 to take with you. You expect to spend $15 each day. You want to have $30 remaining at t

he end of the vacation. Write an explicit formula to represent this scenario. For how many days can you spend $15 a day?
Mathematics
1 answer:
JulijaS [17]3 years ago
6 0

Answer:

105-15x=30

5 days

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R = wp
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3 years ago
i need help with number 2 explain please!! no links and random stuff such as: nooooooo or ahdwhfhaifwiqof.
Vinil7 [7]
I think:
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3 0
3 years ago
Aidan has one box that is 3 3/11 feet tall and a second box that is 3.27 feet tall if he stacks of boxes how tall will the stack
kow [346]

Let's solve this using decimal places.

convert 3 3/11 to decimal number by dividing 3 by 11 and adding it to 3.

3.27 (this is enough places since the other number has 2 places)

Add 3.27 f to 3.27 f.

6.54 feet

3 0
3 years ago
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)

7 0
3 years ago
A student walks 1/4 Mile from her home to the store on her way to a friend's house if the store is 1/3 of the way to a friend's
exis [7]

the store is 1/3 the way to her friend's house, not 1/3 mile.

this explains solution:

1/3x = 1/4

x = 1/4 * 1/3

x = 3/4

Notes:

X: is the total distance.

1/3 is the one portion of total distance.

1/4 mile is the distance store.

3 0
4 years ago
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