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3 years ago

Please help me! It’s timed!!!!!

Mathematics

1 answer:

drek231 [11]3 years ago

8 0

The second option
I think, sorry if you get it wrong

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Show your work. -2.5(4x -4) = -6

serg [7]

Answer:

x = 1 3/5

Step-by-step explanation:

distribute.

-2.5*4x + (-2.5*-4) = -6

-10x +10 = -6

subtract 10 on both sides

-10x = -6 -10

divide by -10

x = -16/-10

x = 1 3/5

7 0

2 years ago

Consider the following two ordered bases of R3:

grigory [225]

Answer:

Let $A = (a_1, ..., a_n)$ and $B = (b_1, ..., b_n)$ bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors $b_1, ..., b_n$ at base A.

The, we case correspond to find the coordinates of vectors of C,

$\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] \}$

at base B.

1. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right]$

now we use backward substitution

$c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\2\\1\end{array}\right]$

2. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right]$

now we use backward substitution $c=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\3\\2\end{array}\right]$

3. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&-3\\-1&2&-1&1\\0&-1&1&2\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&-3\\0&-1&1&2\\0&0&1&-2\end{array}\right]$

now we use backward substitution $c=-2\\-b+c=2,\; b=-4\\a-2b+2c=2,\; a=-2$

Then the coordinate vector of $\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]$ is $\left[\begin{array}{ccc}-2\\-4\\-2\end{array}\right]$

Then the change of basis matrix from B to C is

$\left[\begin{array}{ccc}4&4&-2\\2&3&-4\\1&2&-2\end{array}\right]$

4 0

4 years ago

The height (feet) of an object moving vertically is given by s= (-16t^2)+(208t)+(156), where t is in seconds.

OverLord2011 [107]

Answer:

The object's velocity at t = 7 is -16 ft/s

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

Step-by-step explanation:

From the information given, the equation of motion of the object is

$s(t)= -16t^2+208t+156$

For any equation of motion s(t), the instantaneous velocity at time t is given by

$v(t)=\frac{ds}{dt}$

(a) To find the object's velocity at t = 7, you must:

$v(t)=\frac{d}{dt}(-16t^2+208t+156)\\\\v(t)=-\frac{d}{dt}\left(16t^2\right)+\frac{d}{dt}\left(208t\right)+\frac{d}{dt}\left(156\right)\\\\v(t)=-32t+208+0\\\\v(t)=-32t+208$

Next, we evaluate when t = 7

$v(7)=-32(7)+208=-224+208\\\\v(7)=-16$

The object's velocity at t = 7 is -16 ft/s

(b) <em>To find the maximum of a function we always use the derivative of the function and we set it equal zero to find the</em><em> critical points.</em>

To find the maximum height and when it occurs, we set the velocity function equal to 0 and solve for t.

$-32t+208=0\\\\-32t+208-208=0-208\\\\-32t=-208\\\\\frac{-32t}{-32}=\frac{-208}{-32}\\\\t=\frac{13}{2}$