The answer is 23 because they have to add up to 90 :) - beanz
To solve the problem shown above you must apply the following proccedure:
1. You have the following function given in the problem:
<span> f(x)=x3–3x–2
2. When you give values to the x and plot each point obtained, you obtain the graph shown in the figure attached.
3. Based on the graph and analizing the alternate form:
f(x)=(x-2)(x+1)</span>²
As you can see, there is two roots x=-1, then, you can conclude that the correct answer is the option b, which is:
b) -1<span>
</span>
Y = -2/3x + 7.....the slope here is -2/3. A perpendicular line will have a negative reciprocal slope. All tht means is take the original slope, flip it, and change the sign. So we take -2/3....flip it making it -3/2.....change the sign making it 3/2. So ur perpendicular line will have a slope of 3/2.
y = mx + b
slope(m) = 3/2
(-5,6)....x = -5 and y = 6
now sub and find b, the y int
6 = 3/2(-5) + b
6 = -15/2 + b
6 + 15/2 = b
12/2 + 15/2 = b
27/2 = b
so ur perpendicular equation is : y = 3/2x + 27/2 <==
519/6=86.5
915/7=130.7142
439/7=62.7142
812/9=90.2222
so i think its the 1st on but i may be wrong.
<span>, y+2 = (x^2/2) - 2sin(y)
so we are taking the derivative y in respect to x so we have
dy/dx use chain rule on y
so y' = 2x/2 - 2cos(y)*y'
</span><span>Now rearrange it to solve for y'
y' = 2x/2 - 2cos(y)*y'
0 = x - 2cos(y)y' - y'
- x = 2cos(y)y' - y'
-x = y'(2cos(y) - 1)
-x/(2cos(y) - 1) = y'
</span><span>we know when f(2) = 0 so thus y = 0
so when
f'(2) = -2/(2cos(0)-1)
</span><span>2/2 = 1
</span><span>f'(2) = -2/(2cos(0)-1)
cos(0) = 1
thus
f'(2) = -2/(2(1)-1)
= -2/-1
= 2
f'(2) = 2
</span>