<h3>
Answer: 5</h3>
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Explanation:
Vertex form is
y = a(x-h)^2 + k
We are told the vertex is (3,-2), so we know (h,k) = (3,-2)
y = a(x-h)^2 + k will update to y = a(x-3)^2 - 2
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Then we also know that (x,y) = (4,3) is a point on the parabola. Plug those x and y values into the equation and solve for 'a'
y = a(x-3)^2 - 2
3 = a(4-3)^2 - 2
3 = a(1)^2 - 2
3 = a - 2
3+2 = a
5 = a
a = 5
This is the coefficient of the x^2 term since the standard form is y = ax^2+bx+c.
Answer:
it would be the first answer
Step-by-step explanation:
you can draw the net of the shapes, and count each corner there
ok you need formula for surface area.
SA = πrh+2πr^2
Plug in 25 and 5
SA = 2π(5)(25)+2π(5)^2
SA = 10pi*25 + 2pi(25)
SA = 250pi + 50pi
SA = 300pi
Hope this helps :)
Jeron
Alejandro is correct because you do not need to add another number to 76.518, as Jiro did, but you need to round it to the nearest tenth.
Glad I could help, and good luck!
<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
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