14,28,42,56,70,84,98,112,126,140,154,168,182,196,210,224,238,252,266,280,294,308,
Is that enough?
Answer:
A 1:6
Step-by-step explanation:
3:18 simply to 1:6
Answer:
0.50 kg of the material would be left after 10 days.
0.25 kg of the material would be left after 20 days.
Step-by-step explanation:
We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after 10 days and 20 days, respectively.
We will use half life formula.
, where,
A = Amount left after t units of time,
a = Initial amount,
t = Time,
h = Half-life.
![A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}](https://tex.z-dn.net/?f=A%3D1%5Ctext%7B%20kg%7D%5Ccdot%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Cfrac%7B10%7D%7B10%7D%7D)
![A=1\text{ kg}\cdot(\frac{1}{2})^{1}](https://tex.z-dn.net/?f=A%3D1%5Ctext%7B%20kg%7D%5Ccdot%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B1%7D)
![A=1\text{ kg}\cdot\frac{1}{2}](https://tex.z-dn.net/?f=A%3D1%5Ctext%7B%20kg%7D%5Ccdot%5Cfrac%7B1%7D%7B2%7D)
![A=0.5\text{ kg}](https://tex.z-dn.net/?f=A%3D0.5%5Ctext%7B%20kg%7D)
Therefore, amount of the material left after 10 days would be 0.5 kg.
![A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}](https://tex.z-dn.net/?f=A%3D1%5Ctext%7B%20kg%7D%5Ccdot%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Cfrac%7B20%7D%7B10%7D%7D)
![A=1\text{ kg}\cdot(\frac{1}{2})^{2}](https://tex.z-dn.net/?f=A%3D1%5Ctext%7B%20kg%7D%5Ccdot%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B2%7D)
![A=1\text{ kg}\cdot\frac{1^2}{2^2}](https://tex.z-dn.net/?f=A%3D1%5Ctext%7B%20kg%7D%5Ccdot%5Cfrac%7B1%5E2%7D%7B2%5E2%7D)
![A=1\text{ kg}\cdot\frac{1}{4}](https://tex.z-dn.net/?f=A%3D1%5Ctext%7B%20kg%7D%5Ccdot%5Cfrac%7B1%7D%7B4%7D)
![A=0.25\text{ kg}](https://tex.z-dn.net/?f=A%3D0.25%5Ctext%7B%20kg%7D)
Therefore, amount of the material left after 20 days would be 0.25 kg.
Answer:
a. when x=1 so substitute 1 to any X in the given equation.
y=-3(1)+2.5
y=-3+2.5
y=-0.5
b.when x is -1.5 so substitute -1.5 to any x in the equation
y=-3(-1.5)+2.5
y=-4.5+2.5
y=-2