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Sedbober [7]
3 years ago
13

A curve passes through the point 1,-11 and its gradient at any point is ax2+b, where a and b are constants. The tangent to the c

urve at point 2,-16 is parallel to the x axis. Find the values of a and b.
Mathematics
1 answer:
Blizzard [7]3 years ago
5 0

Answer:

a = 3

b = -12

Step-by-step explanation:

gradient function (y') = ax² + b

curve function (y) = 1/3 ax³ + bx + c  

pass (1,-11) : -11 = 1/3 a + b + c  ..... (1)

pass (2,-16) : -16 = 8/3 a + 2b + c  ..... (2)

slope at (2,-16) is 0 (its tangent parallel to x axis)

at (2 , -16) ax² + b = 0         4a + b = 0         b = -4a   .... (3)

(2) - (1) : -5 = 7/3 a + b = 7/3 a -4a = -5/3 a   ... (4)

3 x (4) : -15 = -5a            a = 3

b = -4a = -4 x 3 = -12

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3 years ago
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tia_tia [17]
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3 years ago
D3a2−(c+b) if a=−2, b=3, c=−12, and d=−4
UkoKoshka [18]

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57

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8 0
3 years ago
Find the area of the shaded region
Alex777 [14]

∆BOC is equilateral, since both OC and OB are radii of the circle with length 4 cm. Then the angle subtended by the minor arc BC has measure 60°. (Note that OA is also a radius.) AB is a diameter of the circle, so the arc AB subtends an angle measuring 180°. This means the minor arc AC measures 120°.

Since ∆BOC is equilateral, its area is √3/4 (4 cm)² = 4√3 cm². The area of the sector containing ∆BOC is 60/360 = 1/6 the total area of the circle, or π/6 (4 cm)² = 8π/3 cm². Then the area of the shaded segment adjacent to ∆BOC is (8π/3 - 4√3) cm².

∆AOC is isosceles, with vertex angle measuring 120°, so the other two angles measure (180° - 120°)/2 = 30°. Using trigonometry, we find

\sin(30^\circ) = \dfrac{h}{4\,\rm cm} \implies h= 2\,\rm cm

where h is the length of the altitude originating from vertex O, and so

\left(\dfrac b2\right)^2 + h^2 = (4\,\mathrm{cm})^2 \implies b = 4\sqrt3 \,\rm cm

where b is the length of the base AC. Hence the area of ∆AOC is 1/2 (2 cm) (4√3 cm) = 4√3 cm². The area of the sector containing ∆AOC is 120/360 = 1/3 of the total area of the circle, or π/3 (4 cm)² = 16π/3 cm². Then the area of the other shaded segment is (16π/3 - 4√3) cm².

So, the total area of the shaded region is

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7 0
2 years ago
Help me please and thank you!!!
loris [4]

Answer:

f(3) = - 60

Step-by-step explanation:

To evaluate f(3) substitute x = 3 into f(x), that is

f(3) = 5(3)² - 7(4(3) + 3)

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4 0
3 years ago
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