The distance from the center of dilation, P, to.the image vertice S' is; 6 units.
<h3>What is the distance from the center of dilation, P, to the image S'?</h3>
It follows from the task content that the center of dilation of the triangle QRS is point P and the length of segment PS in the pre-image is; 8 units.
Hence, since the dilation factor as given in the task content is; three-fourths, it therefore follows that the distance of point P to S' in the image is; (3/4) × 8 = 6units.
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<h3>
Answer: p = 0.12*(1.02)^t</h3>
Explanation:
The general exponential growth equation is
p = A*B^t
where t is the number of years that have gone by after 2008, A is the starting amount, B is the growth multiplier, and p is the price t years after 2008
We know that A = 0.12 is the starting price
The value of B is B = 1.02 which is in the form 1+r since 1.02 = 1 + 0.02 = 1+r
The r value is r = 0.02 and it is positive to represent growth. Keep in mind that 2% = 2/100 = 0.02
So we go from
p = A*B^t
to
p = 0.12*(1.02)^t
Answer:
Answer is 6.23u because ,
Answer:
This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3
Step-by-step explanation:
The given function is
When we differentiate this function with respect to x, we get;
We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)
This implies that;
![c-3=\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c-3%3D%5Csqrt%5B3%5D%7B63.15789%7D)
![c=3+\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c%3D3%2B%5Csqrt%5B3%5D%7B63.15789%7D)
If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).
But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.
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