All categories

3 years ago

Can you find the distance between two points using absolute value

Mathematics

2 answers:

kenny6666 [7]3 years ago

7 0

Answer: yes you can

Step-by-step explanation:

EleoNora [17]3 years ago

3 0

This depends. The way you do this is by taking these two points and by adding their absolute values together, this therefore finds their difference. This only works if one number is positive and one in negative. If two numbers are on the same side of the number line you CANNOT find their difference with absolute value.

Hope this helps, if not, comment below please!!!!

You might be interested in

2. f(x) = 3(x - 2)2 + 6<br>a. Standard Form:

Alexandra [31]

Answer:

${3x}^{2} - 4x + 10 = 0$

Step-by-step explanation:

$3(x - 2) {}^{2} + 6 \\ 3 \times x {}^{2} - 4x + {2}^{2} + 6 \\ 3x {}^{2} - 4x + 4 + 6 \\ {3x}^{2} - 4x + 10 = 0$

6 0

3 years ago

write a polynomial function of least degree with integral coefficients having zeros that include -1 and 1 + 2i

SOVA2 [1]

Answer:

Step-by-step explanation:

hello :

a polynomial function is:

f(x) = a (x+1)(x-1)(x-2i) .... a ≠ 0

7 0

3 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is: