0.0505(35) = 1.7675...rounds to 1.77
157 - 35 = 122
0.0305(122) = 3.721...rounds to 3.72
the fee is : 1.77 + 3.72 = $5.49
Answer:
The function f(x) is not given, I used a different function but the approach and steps is the same .
Step-by-step explanation:
- Given the function f(x) = 2x2 - 8x + 5
compare with the normal quadratic equation ; ax2 + bx + c = f(x)
- since a is greater than zero i.e a > o {positive}
As such, it has a minimum
hence for minimum value; x = -b/2a
x = -(-8)/2 x 2
x = 8/4 = 2
plugging the values of x in f(x) ; f(2) = 2(2)^2 -8(2) + 5
f(2) = -3, hence it has minimum value and the minimum value is -3
<h2>Answer-Average rate of change(A(x)) of f(x) over a interval [a,b] is given by:</h2><h2 /><h2>A(x) = \frac{f(b)-f(a)}{b-a}A(x)= </h2><h2>b−a</h2><h2>f(b)−f(a)</h2><h2> </h2><h2> </h2><h2 /><h2>Given the function:</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^xf(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>x</h2><h2> </h2><h2 /><h2>We have to find the average rate of change from x = 1 to x= 2</h2><h2 /><h2>At x = 1</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^1 = 5f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>1</h2><h2> =5</h2><h2 /><h2>At x = 2</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^2=20 \cdot \frac{1}{16} = 1.25f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>2</h2><h2> =20⋅ </h2><h2>16</h2><h2>1</h2><h2> </h2><h2> =1.25</h2><h2 /><h2>Substitute these in above formula we have;</h2><h2 /><h2>A(x) = \frac{f(2)-f(1)}{2-1}A(x)= </h2><h2>2−1</h2><h2>f(2)−f(1)</h2><h2> </h2><h2> </h2><h2 /><h2>⇒A(x) = \frac{1.25-5}{1}=-3.75A(x)= </h2><h2>1</h2><h2>1.25−5</h2><h2> </h2><h2> =−3.75</h2><h2 /><h2>therefore, average rate of change of the function f(x) from x = 1 to x = 2 is, -3.75</h2>
<h2>Please Mark me as brainlist. </h2>
Answer:
C) (square root 2)/2
Step-by-step explanation:
In quadrant III, the sine of an angle is positive. The value will be ...
sin(Θ) = √(1 -cos(x)^2) = √(1 -(-√2/2)^2) = √(2/4)
sin(Θ) = (√2)/2
