All categories

2 years ago

Help Please.

Mathematics

2 answers:

gregori [183]2 years ago

8 0

Answer:

2.125ft

Step-by-step explanation:

10×1=10

10×1.25= 12.5

12.5+10= 22.5in

22.5 in = 1.875ft

4-1.875= 2.125ft left

OleMash [197]2 years ago

3 0

Answer:

Step-by-step explanation:

I foot = 12 inches

1 inch = 1/12 foot

If the books are 1.-inch wide, converting to foot, it becomes 1/12 foot wide. It means that the total space occupied by 10 books of this size is

10 × 1/12 = 10/12 = 0.833 foot.

If the books are 1.25-inch wide, converting to foot, it becomes 1/12 × 1.25 = foot wide. It means that the total space occupied by 10 books of this size is

10 × 1/12 × 1.25 = 12.5/12 = 1.042 foot.

Total space occupied by the books would be

0.833 + 1.042 = 1.875 feet

Amount of room left in the bookshelf would be

4 - 1.875 = 2.125 feet

You might be interested in

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:

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