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emmainna [20.7K]
1 year ago
12

Evaluate dydx at x=2 for the function below. y=6u^3+5u+2, where u=−3x^2−4x+9

Mathematics
1 answer:
bekas [8.4K]1 year ago
6 0

Step-by-step explanation:

y = 6u {}^{3}  + 5u + 2

y = 6( - 3 {x}^{2}  - 4x + 9) {}^{3}  + 5( - 3 {x}^{2}  - 4x + 9) + 2

Use the chain rule.

\frac{dy}{dx}  = 6(3)( - 3 {x}^{2}  - 4x + 9) {}^{2} ( - 6x  -  4) + 5( - 6x - 4)

18( - 3 {x}^{2}  - 4x + 9) {}^{2} ( - 6x - 4) + 5( - 6x - 4)

Factor out -6x-4,

( - 6x - 4)(18( - 3 {x}^{2}  - 4x + 9) {}^{2}  + 5)

Know plug in x=-2 to evaluate the derivative

( - 6(2) - 4)(18( - 3(2) {}^{2}  - 4(2) + 9) {}^{2}  + 5)

( - 16)(18(( - 12) - 8 + 9)) {}^{2}  + 5)

- 16(18( - 11) {}^{2}  + 5)

- 16(2183)

- 34928

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I need these 2 questions answered please.​
Sav [38]

The area of the circle is 28 ft square.

The area of the triangle is 30 ft square.

Step-by-step explanation:

1) The area of the circle= π*r^2

where π= 3.14(default value) and "r" is the radius of the circle.

Here, Radius of the circle(r)= 3 ft

Area = 3.14*3*3

        = 28.26 ft sq.= 28 ft sq(rounded to the nearest whole number)

2) The triangle has the base= 10 feet and height= 6 feet

The Area of the triangle= 1/2(b)(h)

where "b" is the base and "h" is the height of the triangle.

Substitute b=10 and h=6 ,

Area of triangle= 1/2(10 ft)(6 ft)= 60/2= 30 feet sq.

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4 years ago
I NEED HELP IMMEDIATELY PLEASE AND THANK YOU!<br> Solve 7−√159−2=−5
Yuri [45]

Answer:

7-12.6-2=-5

-7.6=-5

7.6-5

2.6

4 0
3 years ago
Wats the answer for this math or
allochka39001 [22]
The answer is D because the whole circle is 369 degrees the big arc is 202 so 360- 202 is 158 then you have an inscribed angle of 35 multiply it by two to find the arc of that so it would be 70 so subtract 158-70=88 and you can see that that section of 88 is part of an inscribed angle so divide by two to get the answer of 44
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3 years ago
Solve the quadratic equation by completing the square.
Step2247 [10]

Answer:

x1, x2 = 4.74 , -2.74

Step-by-step explanation:

To find the roots of a quadratic function we have to use the bhaskara formula

ax^2 + bx + c

x^2 - 2x - 13

a = 1     b = -2    c = -13

x1 = (-b + √ b^2 - 4ac)/2a

x2 =(-b - √ b^2 - 4ac)/2a

x1 = (2 + √ (2^2 - 4 * 1 * (-13)))/2 * 1

x1 = (2 + √ (4 + 52)) / 2

x1 = (2 + √ 56 ) / 2

x1 = (2 + 7.48) / 2

x1 = 9.48 / 2

x1 = 4.74

x2 = (2 - √ (2^2 - 4 * 1 * (-13)))/2 * 1

x2 = (2 - √ (4 + 52)) / 2

x2 = (2 - √ 56 ) / 2

x2 = (2 - 7.48) / 2

x2 = -5.48 / 2

x2 = -2.74

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