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2 years ago

Help me anyone ASAP snens

Mathematics

1 answer:

azamat2 years ago

5 0

Answer:

thousandths

Step-by-step explanation:

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If I simplify the expression 4(x+3) to 4x+7 what properties did I use

natka813 [3]

You used Order of Operations. I hope this helped ^^

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3 years ago

Please help solve this!

algol13

Answer:

There’s nothing to solve?

Step-by-step explanation:

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2 years ago

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Sorry my brain is fried...

KiRa [710]

Answer:

1:D,2

2:B:She found that -3(9) is positive

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4:Confused, is there supposed to be supplemental info?

Step-by-step explanation:

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2 years ago

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Need help ASAP no Rocky

Ahat [919]

Answer:

The length of mid-segment $XY=4$ units

Step-by-step explanation:

Given:

Length of $QR=8$ units

$XY$ is the mid-segment which means the line joining the midpoints of two sides of triangle.

By mid-segment theorem the mid-segment of a triangle is parallel to the third side and half of the length of third side.

So, we can write:

$XY=\frac{1}{2}QR$

$XY=\frac{1}{2}(8)$

∴ $XY=4$ units

∴The length of mid-segment $XY=4$ units

5 0

2 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

2 years ago