Answer and Step-by-step explanation:
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1. Let's write out the equation to subtract.
7t - 2u - 3v - (t - 3v)
Distribute the negative to the t and -3v.
7t - 2u - 3v - t + 3v (The negatives cancel out)
Now simplify by combining like terms.
6t - 2u
This is the answer because the 3v and -3v cancel out.
2. I don't really understand what this is saying. Is there answer choices for this? But what I think its saying is that the lift has a constant of 2.
3. To find out the amount of terms, we would simplify the equation.
2x + 3y - 5x + yz - x
-4x + 3y + yx
Here, we can see that we have 3 terms in this expression.
-4x is the first term, +3y is the second term, and +yx is the third term.
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#teamtrees #WAP (Water And Plant)
Answer: OPTION A.
Step-by-step explanation:
You can observe that in the figure CDEF the vertices are:

And in the figure C'D'E'F' the vertices are:

For this case, you can divide any coordinate of any vertex of the figure C'D'E'F' by any coordinate of any vertex of the figure CDEF:
For C'(-8,-4) and C(-2,-1):

Let's choose another vertex. For E'(8,8) and E(2,2):

You can observe that the coordinates of C' are obtained by multiplying each coordinate of C by 4 and the the coordinates of E' are also obtained by multiplying each coordinate of E by 4.
Therefore, the rule that yields the dilation of the figure CDEF centered at the origin is:
→
Answer:
5
Step-by-step explanation:
Use pythagorean theorem. The rise is 3 units and the run is 4 unites. Now the length of the line we are trying to find is a hypothenuse(the side opposite of the right angle
If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y = 
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 = 
Cancel the C in both side
1/2 = 
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 × -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
Learn more about half life here
brainly.com/question/4318844
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