All categories

2 years ago

Please help me find the value of this angle?

Mathematics

1 answer:

yawa3891 [41]2 years ago

7 0

I'm pretty sure it would have to be 155. Considering a flat line has an angle of 180 and if you subtract 180 from 25 you get 155.

You might be interested in

The slope of the line y=2/3x+5/9 is M=

beks73 [17]

Answer: 7

Step-by-step explanation:

8 0

2 years ago

When g ( t ) = 3 t 2 + 2 t + 1 , evaluate g ( − 4 )

Luda [366]

Answer:

Just substitue every 't' with -4

Step-by-step explanation:

g(-4) = 3*(-4)*(-4) + 2*(-4) + 1

g(-4) = 64-8+1

g(-4) = 57

7 0

2 years ago

Read 2 more answers

Write a system of linear equations for the graph below.

Vanyuwa [196]

Answer:

-3/-2

Step-by-step explanation:

because (x,y) the x coordinate is -3 and the y coordinate is -2

3 0

2 years ago

Rewrite (3b) to the second power without the exponents

Alik [6]

3bx3bx3b

27xbxbxb

hope this helps

5 0

2 years ago

the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a

Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - First derivative of mass in time, measured in miligrams per year.

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

$m$ - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

$\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt$

$\ln m = -\frac{1}{\tau} + C$

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$m_{o}$ - Initial mass of isotope, measured in miligrams.

$t$ - Time, measured in years.

And time is cleared within the equation:

$t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]$

Then, time constant can be found as a function of half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

If we know that $t_{1/2} = 5730\,yr$ and $\frac{m(t)}{m_{o}} = 0.35$ , then:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.35$

$t \approx 8678.505\,yr$

The wood was cut approximately 8679 years ago.

5 0

2 years ago