6m^3+47m^2-16m-12/3m-2

The length of the garden is 10 feet more than twice its width. The perimeter of the garden is 170 feet.

kumpel [21]

Answer:

Width is 25 ft

length is 60 ft

Step-by-step explanation:

Here, we want to calculate the length and width of the garden

From the question, we are told that the length is 10 ft more than twice its width

So if the width is x ft

Length = 2x + 10

Mathematically, the perimeter of a rectangle i s 2(l + b)

So;

2(x + 2x + 10) = 170

2(3x + 10) = 170

3x + 10 = 85

3x = 85-10

3x = 75

x = 75/3

x = 25 ft

length is;

2(25) + 10 = 50 + 10 = 60 ft

8 0

3 years ago

The cost of every college is expected to increase 3.5%

lions [1.4K]

Answer:

$18,630

Step-by-step explanation:

The cost of every college is expected to increase 3.5% next year.

The cost to attend Westish College is currently $18000.

What is the expected cost to attend Westish College next year?

The increase in the cost for next year is calculated as:

3.5 % × Current cost

= 3.5% × $18,000

= $630.

The expected cost to attend Westish college next year=

$18,000 + $630

= $18,630

8 0

3 years ago

Which of the following could be the ratio between the lengths of the two legs

skad [1K]

9514 1404 393

Answer:

B, E

Step-by-step explanation:

The ratios of side lengths in a 30-60-90 triangle are ...

1 : √3 : 2

The two legs are the shorter sides, so any ratio that reduces to 1 : √3 is an appropriate choice:

B. 1 : √3

E. 2√3 : 6

3 0

3 years ago

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

4 0

3 years ago

A number is greater than or equal to -4<br>

Trava [24]

It depends on the number

6 0

2 years ago

Read 2 more answers