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3 years ago

HELP PLEASE!

Mathematics

1 answer:

Jobisdone [24]3 years ago

3 0

Answer:

24 in long by 12 in wide.

Step-by-step explanation:

l=2w

288=(2w)×w

288=2w^2

144=w^2

12=w

use w=12 to solve for l

l=2(12)

l=24

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A map of the United States has a scale of 1 inches = 300 miles. The distance on the map from Boise, Idaho, to Phoenix, Arizona,

Nostrana [21]

Answer:

The distance is 750 miles.

Step-by-step explanation:

In order to find this, we create a proportion. On the left would be the scale factor. On the right would be the Idaho to Arizona distance.

1in / 300 mi = 2.5in / x mi

Now we can cross multiply to find the distance.

300 * 2.5 = 1 * x

750 mi = x

4 0

4 years ago

Which of the following points is on the graph of….

bonufazy [111]

we have the equation

$y=\log_{\frac{1}{2}}x$

Remember that

If an ordered pair is on the graph of the given function, then the ordered pair must satisfy the given function

Verify each option

N 1

(0,1)

x=0, y=1

Is not true, because the log of zero is undefined

N 2

(1,1/2)

x=1, y=1/2

Is not true, because the log of 1 is zero

N 3

(1/2,1)

$y=\operatorname{\log}_{\frac{1}{2}}\frac{1}{2}=1$

Is true

therefore

<h2>The answer is (1/2,1)</h2>

8 0

1 year 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

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