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2 years ago

The expression describe the number that is 8 to the left of -8 on the number line

Mathematics

1 answer:

mart [117]2 years ago

4 0

Start at -8 move to the left 8 more times,. You get -16.

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erry jogs the same round trip path everyday. He jogs 0.7 miles to the park and then 0.5 miles from the park to the playground. A

adelina 88 [10]

To solve this problem you must use the information given in the problem above:

1. You have that he jogs $0.7miles$ to the park and then $0.5miles$ to the playground, then he jogs $0.3$ miles to the bridge and $0.9miles$ to his house.

2. Therefore, as you can see, Jerry completes his jog when he arrives to his house. So, he is $0miles$ from his house.

Then, the answer is: $0miles$

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2 years ago

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Which two ratios represent quantities that are proportional?

Ad libitum [116K]

Answer:

All 4 of them

Step-by-step explanation:

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2 years ago

X=2t/t^2 +1 and y=2t/t^2 +1 <br>Find dy/dx

zhenek [66]

Answer:

dy/dx=1

Hope this helps! Please vote brainliest! :)

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2 years ago

George said if 2 and 4 are factors of a number, then 8 is a factor of the whole number. is he correct explain

Mamont248 [21]

He is correct and incorrect. 2 and 4 can create factors but they can also be factors of other numbers. So 8 can be a factor of bigger whole numbers and can also have it own smaller factors.

8 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

The expression describe the number that is 8 to the left of -8 on the number line ​

The expression describe the number that is 8 to the left of -8 on the number line