5r^2-3rs<br> r=4<br> s=5 <br> help

Two airliners are 1600 miles apart and heading toward each other at different altitudes. The first plane is traveling north at 6

viktelen [127]

In approximately around 2 hours and 10 mins

4 0

3 years ago

Write 23 hundredths in scientific notation

Dmitry [639]

Answer:

Step-by-step explanation:

23 hundredths = 0.23 = 2.3 * 10⁻¹

8 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

3 Prove that<br>cos130°+cos 110+ cos10=0

Evgesh-ka [11]

We know, cos A + cos B = 2cos(C+D)/2 × cos(C - D)/2

So,

cos 130 + cos 110 + cos 10

cos 130 + 2cos (110+10)/2 × cos( 110-10)/2

cos 130 + 2cos 60 × cos 50

cos ( 180 - 50 ) + 2cos60 cos 50

We know, cos ( 180 - A ) = -cos A

2cos 60 cos 50 - cos 50

Also, cos 60 = 0.5

2×0.5 cos 50 - cos50

cos 50 - cos 50

0 = RHS.

Hence proved

5 0

3 years ago

I am thinking of a 3-digit number.

koban [17]

Answer:

The answer is 939

Step-by-step explanation:

If you divide 939 by 9 it equals 104 R3.

If you divided 939 by 2 it equals 469 R1

And if you divide 939 by 5 it equals 187 R4

Hope this helps!

7 0

3 years ago

5r^2-3rs r=4 s=5 help