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poizon [28]
3 years ago
14

Steve has 50 notebooks with red covers but only 40 notebooks with blue covers. If he uses up 6 notebooks with red covers and 4 w

ith blue covers each month, in how many months will he have the same number of red as blue notebooks?
Mathematics
1 answer:
Akimi4 [234]3 years ago
6 0

Answer:

5 months

Step-by-step explanation:

The first thing to do is to obtain the difference between the two numbers of colored books, 50 and 40.

50 - 40 = 10 notebooks.

We need to calculate how long it will take for this difference to disappear, noting that Steve uses more of the red cover notebooks every month. The difference on a monthly basis is

6 - 4 = 2 notebooks per month

So, now the number of months will be gotten from;

Number of months = \frac{Total Difference}{Monthly Difference}

Number of months = \frac{10}{2} = 5 months

We can check for this,

After 5 months, Steve would have used 5 × 6 red cover notebooks = 30

and 5 × 4 blue cover notebooks = 20

The remaining number of books will be:

RED: 50 - 30 = 20 red cover notebooks

BLUE: 40 - 20 = 20 blue cover notebooks.

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in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)
Nimfa-mama [501]

Answer:

The component of \vec u orthogonal to \vec a is \vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right).

Step-by-step explanation:

Let \vec u and \vec a, from Linear Algebra we get that component of \vec u parallel to \vec a by using this formula:

\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a (Eq. 1)

Where \|\vec a\| is the norm of \vec a, which is equal to \|\vec a\| = \sqrt{\vec a\bullet \vec a}. (Eq. 2)

If we know that \vec u =(2,1,1,2) and \vec a=(4,-4,2,-2), then we get that vector component of \vec u parallel to \vec a is:

\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)

\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)

\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)

Lastly, we find the vector component of \vec u orthogonal to \vec a by applying this vector sum identity:

\vec  u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a} (Eq. 3)

If we get that \vec u =(2,1,1,2) and \vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right), the vector component of \vec u is:

\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10}    \right)

\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)

The component of \vec u orthogonal to \vec a is \vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right).

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Answer:

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