Dree rolls a strike in 6 out of the 10 frames of bowling

Find an equation of the line having the given slope and containing the given point. m=-7,(5,9) the equation of the line is y=

k0ka [10]

Open this web, the web is really - really helps.................................... http://www.coolmath.com/algebra/08-lines/11-finding-equation-line-point-slope-01

6 0

3 years ago

Which mathematical property is demonstrated?

RSB [31]

Answer:

The commutative property states that you can rearrange the order of the numbers and get the same result. The commutative property states that you can rearrange the order of the numbers and get the same result. The addition properties are the exact same, but replace multiply with add. Your answer is D.

Step-by-step explanation:

All the problem is doing is switching the numbers to different sides of the equation. You will still get the same answer.

5 0

3 years ago

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Compare the decimals 14.40 and 14.04.

tatuchka [14]

C. 14.40 is greater than 14.04

8 0

2 years ago

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Vanessa has scored 45 32 and 37 are three math quizzes she will take one more quiz and she wants a quiz average of at least 40 w

dedylja [7]

She must at least score a 38
To get the average you need to add 45 + 32 + 37 together than divide it by the amount of numbers added, in this case 3.

45 + 32 + 37 = 114
114÷3 = 38

3 0

2 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

2 years ago