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Ira Lisetskai [31]

3 years ago

6

2 nonoverlapping acute angles that share a ray form an obtuse angle. If one of the acute angles has a measure of 50 degrees what

could be the measure of the other acute angle?

1 answer:

Ne4ueva [31]3 years ago

4 0

It would be 50 an acute angle is always under 50 atleast

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What is the degree of -7xyz^2

leva [86]

Answer:

The degree is 4

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4 years ago

3.) If y varies directly as x and y = 24 when x = 6, find the variation constant and the equation of variation.

Talja [164]

<h2><u>Problem</u>:-</h2>

3.) If y varies directly as x and y = 24 when x = 6, find the variation constant and the equation of variation.

<h2><u>Solution</u>:-</h2>

A. Express the statement “y varies directly as x”, as y = kx .

B. Solve for k by substituting the given values in the equation.

$\sf\rightarrow{y = kx}$

$\sf\rightarrow{24 = 6k}$

$\sf\rightarrow{K = \frac{24}{6} }$

$\sf\rightarrow{K={\color{magenta}{4}}}$

<h2><u>Answer</u>:-</h2>

Therefore, the constant of variation is 4.

C. Form the equation of the variation by substituting 4 in the statement y = kx. Thus , <u>y = 4 x.</u>

<u> ${\large{—————————————————————}}$ </u>

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3 years ago

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3 years ago

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4 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:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

$\Delta t_{1/2} \approx 65.788\,days$

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

4 0

3 years ago

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