In linear models there is a constant additve rate of change. For example, in the equation y = mx + b, m is the constanta additivie rate of change.
In exponential models there is a constant multiplicative rate of change.
The function of the graph seems of the exponential type, so we can expect a constant multiplicative exponential rate.
We can test that using several pair of points.
The multiplicative rate of change is calcualted in this way:
[f(a) / f(b) ] / (a - b)
Use the points given in the graph: (2, 12.5) , (1, 5) , (0, 2) , (-1, 0.8)
[12.5 / 5] / (2 - 1) = 2.5
[5 / 2] / (1 - 0) = 2.5
[2 / 0.8] / (0 - (-1) ) = 2.5
Then, do doubt, the answer is 2.5
Answer:
2
Step-by-step explanation:
In this case, you have to substitute the w for -1, and then do the operation:
-6( -1 )^2 + 13( -1 ) + 21
-6( 1 ) - 13 +21
-6 -13 + 21
-19 + 21
2
Given:
- The principal amount that Amy opened her savings account with is $1750.
- The rate of simple interest compounded annually is 4.3%.
- The time period for which we calculate the new balance is 6 months.
To Find:
The balance after 6 months.
Answer:
The balance after 6 months will be $1787.625
Step-by-step explanation:
The principal amount that Amy opened her savings account with is $1750. We can denote this by P.
The rate of simple interest compounded annually is 4.3% which we may denote by R.
The time period for which we calculate the new balance is 6 months which can be written as 0.5 years (since the rate of interest is compounded annually, we must consider the time period in terms of years).
The amount of money accrued from the interest can be calculated by the formula

Putting in the values given in the question, we have

The amount in the bank account will be the principal amount plus the amount of interest accrued that we have calculated above.
Thus, the balance after 6 months will be 1750 + 37.625 = $1787.625.
The middle right is 145-90, so 55°. I'm not sure about the others
Try adding the two and think of it mentally as if you were drinking the amount of cups of water that's what I do when I have questions like these :))