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Dennis_Churaev [7]

2 years ago

13

Element X is a radioactive isotope such that every 27 years, its mass decreases by half. Given that the initial mass of a sample

of Element X is 7900 grams, how much of the element would remain after 12 years, to the nearest whole number?

2 answers:

mafiozo [28]2 years ago

6 0

Answer:e

Step-by-step explanation:

Oksi-84 [34.3K]2 years ago

3 0

Answer:

5805

Step-by-step explanation:

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Simplify,<br> Rewrite the expression in the form 2n.<br> 2.2.2.2.2.2.2.2<br> 2.2.2.2

Nastasia [14]

Answer: 8/4

Step-by-step explanation: if I’m wrong just watch the video for a hint it helps

6 0

2 years ago

SOLVE FOR f(-2) MATH HELP PLS

tankabanditka [31]

Answer: 9

Step-by-step explanation: Basically what f(-2) means is -2 = x. So you put in -2 wherever x appears. This would turn the equation to -3(-2) + 3 which goes to 6+3.

4 0

3 years ago

Read 2 more answers

What is the length of the line segment that is graphed from (+2, +5) to (+2, +8)?

Natasha_Volkova [10]

The length of the line segment is 3.

Hope this helps!

6 0

3 years ago

Please help me find the value of each variable in the parallelogram

seropon [69]

Answer:

K=10 H=65

Step-by-step explanation:

To work out K, you need to know that opposite angles are the same, so 5k is the same as 50, 50÷5= 10, so k=10

To work out H, you know that opposite angles are the same so 2h is the same as 130, 130÷2=65 so h=65

4 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