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3 years ago

Which of the following is an appropriate line of fit for the data?

Mathematics

1 answer:

Bond [772]3 years ago

6 0

Answer:

123idbd8dbokeofyccbdid12332

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5TH GRADE MATH GIVING BRAINLIESTT

murzikaleks [220]

Answer:

graph A

Step-by-step explanation:

When the cost is at 4 dollars, the feet is at 1. Hope this helps!

7 0

3 years ago

What is the gcf of 48, 64 and 72

ziro4ka [17]

The greatest common factor is 8 because Prime factor is 2 and 4, Number 48 is 4 and 1 Number 64 is 6 and o Number 72 is 3 and 0 which it would all equal 2^3 so 48 = 2^4 times 3 64 = 2^6 and 72 = 2^3 times 3^2 Hope this helps.

5 0

3 years ago

Read 2 more answers

What is the answer A B C D

Kobotan [32]

Answer:

the answer is C

Step-by-step explanation:

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

Which expression is equivalent to one over five m − 20? (4 points)

exis [7]

Answer:

The equivalent to one over five m − 20 is one over five (m − 100) ⇒ B

Step-by-step explanation:

Let us solve the question

∵ One over five means $\frac{1}{5}$

∴ One over five m - 20 = $\frac{1}{5}$ m - 20

→ By using the distributive property, take one over five as a common factor

from both terms

∴ $\frac{1}{5}$ m - 20 = $\frac{1}{5}$ $(\frac{\frac{1}{5}m}{\frac{1}{5}}$ - $\frac{20}{\frac{1}{5}})$

→ Simplify the bracket

∵ $\frac{1}{5}$ m ÷ $\frac{1}{5}$ = $\frac{1}{5}$ m × 5 = m

∵ 20 ÷ $\frac{1}{5}$ = 20 × 5 = 100

∴ $\frac{1}{5}$ $(\frac{\frac{1}{5}m}{\frac{1}{5}}$ - $\frac{20}{\frac{1}{5}})$ = $\frac{1}{5}$ (m - 100)

∴ $\frac{1}{5}$ m - 20 = $\frac{1}{5}$ (m - 100)

The equivalent to one over five m − 20 is one over five (m − 100)

4 0

4 years ago