Answer:
Enlargement
Step-by-step explanation:
If a scale factor is smaller than 1, it is a reduction, and if it's larger than 1, it's an enlargement.
Answer:
B'(-7 , -2)
Step-by-step explanation:
First we must understand the coordinate-axis, when we want to move a point to the left or right we do it on the x-axis. to move up or down is on the y axis.
now if we move to the left we go to the negative and to the right the positive
as we are going to move to the left we have to subtract the value that he gave us (4) only to the part of x
B(-3 , -2)
-3 - 4 = -7
B'( -7 , -2)
4 is probably the answer. Hope this helps. :)
Answer:
1. 1343 years
2. 9 hours
3. 39 years
Step-by-step explanation:
1. Given, half-life of carbon = 5730 years.
∴ λ = 0.693/half-life of carbon = 0.693/5730 = 0.000121
If N₀ = 100 then N = 85
Formula:- N = N₀*e^(-λt)
∴ 85 = 100 * e^(-0.000121t)
∴㏑(-0.85)=-0.000121t
∴ t = 1343 years
2. Given half-life of aspirin = 12 hours
λ = 0.693/12 = 0.5775
Also N₀ = 100 then 70 will disintegrate and N = 30 will remain disintegrated.
∴ 70 = 100 *e^(-0.05775t)
0.70 = e^(-0.05775t)
㏑(0.70) = -0.05775t
∴ t = 9 hours
3. The population of the birds as as A=A₀*e^(kt)
Given that the population of birds fell from 1400 from 1000, We are asked how much time it will take for the population to drop below 100, let that be x years.
The population is 1400 when f = 0, And it is 1000 when f = 5
We can write the following equation :
1400 = 1000e^(5t).
∴1400/1000 = e^(5k)
∴ k = ㏑(1.4)/5
We need to find x such that 1400/100 = e^(xk)
14 = e^(xk)
∴ x = 39 years
The power of products property states that for number
enclosed in a bracket or parenthesis, if it is raised to a power, it must be
multiplied to the power of the enclosed number no matter how different the base
is. You cannot add it because it is not raised. You can only add it if they
have the same base. But in this problem, you will just multiply it. The breakdown
of the solution to this problem is shown below. So,
<span><span>• (2x⁵y²)³=(21x3x5*3y2*3)
= 6x15y6</span><span>
</span></span>
