Which statement describes the changes that were made to the graph of f to make the graph of g

Help me answer this question :((((

ss7ja [257]

Step-by-step explanation:

I cannot give an exact result but I will try.

What it means by this is to continue the graph up to 2010.

Based on its trajectory, you would simply draw the line on.

If I made a prediction on what it would end up on in 2010, I would say 50 percent.

8 0

2 years ago

Fatima wants you to show her some numbers other than pie that are real but not rational what would you show her?

EastWind [94]

REAL BUT NOT RATIONAL :
square root of 2
the golden ratio : 1.618033.....non terminating
square root of 99

8 0

3 years ago

Is 4(5a-8)+2a equivalent to 12a-6

dem82 [27]

No, it would end up being 20a-32 which is not equivalent to 12a-6

8 0

3 years ago

Read 2 more answers

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

I need help with this problem! I'll give brainlist!

aliya0001 [1]

Step-by-step explanation:

use CAH method (Cos angle = adjacent/hypotenuse)

Given adjacent = 5

Hypotenuse = 12,

$\cos(x) = \frac{adjacent}{hypotenuse} \\ \cos(x) = \frac{5}{12} \\ x = {cos}^{ - 1} ( \frac{5}{12} ) \\ = 65.4deg(nearest \: tenth)$

4 0

3 years ago