Answer:
the answer forWhich angle measures are correct? Select three options. mAngle2 = 125° mAngleAngle3 = 55° mAngleAngle8= 55° mAngleAngle12 = 100° mAngleAngle14 = 100°. is A,C,E
Answer:
1 1/4
Step-by-step explanation:
For the strawberries, we will have to multiply 3 by 1/4 to get the total amount of strawberries used.
1/4 * 3 = 3/4
For the blueberries, we will have to multiply 2 by 1/4 to get the total amount of blueberries used.
1/4 * 2 = 2/4
Simplify that to get 1/2
Now we need to add 3/4 and 1/2
3/4 + 1/2 = 5/4
Simplify that and we get our answer;
1 1/4
False. A parallelogram is always a quadrilateral.
Answer:
216 pints or 27 gallons
Step-by-step explanation:
Basically you have to convert the gallons into pints to make it easier to add.
1 gallon = 8 pints
To turn 24 gallons into pints, multiply 24*8= 192.
So, now you just add 24+192 which is 216.
To convert into gallons, divide by 8 so it is 27.
216 pints or 27 gallons
Hope this helps!!
-Ketifa
(Can I get brainliest?)
First all, the decay formula is
![P(t)=Ae^{-kt}](https://tex.z-dn.net/?f=P%28t%29%3DAe%5E%7B-kt%7D%20)
where:
![P(t)](https://tex.z-dn.net/?f=P%28t%29)
is the remaining quantity after
![t](https://tex.z-dn.net/?f=t)
years
![A](https://tex.z-dn.net/?f=A)
is the initial sample
![t](https://tex.z-dn.net/?f=t)
is the time in years
![k](https://tex.z-dn.net/?f=k)
is the decay constant
From the problem we know that
![A=33](https://tex.z-dn.net/?f=A%3D33)
and
![P=100](https://tex.z-dn.net/?f=P%3D100)
, but we don't have the time
![t](https://tex.z-dn.net/?f=t)
; to find it we will take advantage of the half-life of the Carbon-14. If you have a sample of 100 mg and Carbon-14 has a half-life of 5730, after 5730 years you will have half of your original sample i.e. 50 mg. We also know that after
![t](https://tex.z-dn.net/?f=t)
years we have a remaining sample of 33mg, so the amount of the sample that decayed is
![100mg-33mg=67mg](https://tex.z-dn.net/?f=100mg-33mg%3D67mg)
. Knowing all of this we can set up a rule 3 and solve it to find
![t](https://tex.z-dn.net/?f=t)
:
![\frac{100mg--\ \textgreater \ 5730years--\ \textgreater \ 50mg}{100mg--\ \textgreater \ (t)years---\ \textgreater \ 67mg}](https://tex.z-dn.net/?f=%20%5Cfrac%7B100mg--%5C%20%5Ctextgreater%20%5C%205730years--%5C%20%5Ctextgreater%20%5C%2050mg%7D%7B100mg--%5C%20%5Ctextgreater%20%5C%20%28t%29years---%5C%20%5Ctextgreater%20%5C%2067mg%7D%20)
![\frac{5730years--\ \textgreater \ 50mg}{(t)years---\ \textgreater \ 67mg}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5730years--%5C%20%5Ctextgreater%20%5C%2050mg%7D%7B%28t%29years---%5C%20%5Ctextgreater%20%5C%2067mg%7D%20)
![\frac{5730}{t} = \frac{50}{67}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5730%7D%7Bt%7D%20%3D%20%5Cfrac%7B50%7D%7B67%7D%20)
![t= \frac{(5730)(67)}{50}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B%285730%29%2867%29%7D%7B50%7D%20)
![t=7678.2](https://tex.z-dn.net/?f=t%3D7678.2)
Now that we know our time
![t](https://tex.z-dn.net/?f=t)
lets replace all the values into our decay formula:
![33=100e^{-7678.2k}](https://tex.z-dn.net/?f=33%3D100e%5E%7B-7678.2k%7D%20)
Notice that the constant
![k](https://tex.z-dn.net/?f=k)
we need to find is the exponent; we must use logarithms to bring it down, but first lets isolate the exponential expression:
![\frac{33}{100} = e^{-7678.2k}](https://tex.z-dn.net/?f=%20%5Cfrac%7B33%7D%7B100%7D%20%3D%20e%5E%7B-7678.2k%7D%20)
![e^{-7678.2k} = \frac{33}{100}](https://tex.z-dn.net/?f=e%5E%7B-7678.2k%7D%20%3D%20%5Cfrac%7B33%7D%7B100%7D%20)
![ln(e^{-7678.2k} )=ln( \frac{33}{100} )](https://tex.z-dn.net/?f=ln%28e%5E%7B-7678.2k%7D%20%29%3Dln%28%20%5Cfrac%7B33%7D%7B100%7D%20%29)
![-7678.2k=ln( \frac{33}{100} )](https://tex.z-dn.net/?f=-7678.2k%3Dln%28%20%5Cfrac%7B33%7D%7B100%7D%20%29)
![k= \frac{ln( \frac{33}{100}) }{-7678.2}](https://tex.z-dn.net/?f=k%3D%20%5Cfrac%7Bln%28%20%5Cfrac%7B33%7D%7B100%7D%29%20%7D%7B-7678.2%7D%20)
![k=-0.000144](https://tex.z-dn.net/?f=k%3D-0.000144)
We can conclude that the decay constant
![k](https://tex.z-dn.net/?f=k)
is approximately -0.000144