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1 year ago

What is the measure of 1,2, and 3

Mathematics

2 answers:

shutvik [7]1 year ago

8 0

Answer:

1 = 104

2 = 33

3 = 71

Step-by-step explanation:

Angle 1 is 180 - 57- 19 = 104 from supplementary angles

Angle 2 is 90 - 57 = 33 from complementary angles

Angle 3 is 90 - 19 = 71 from complementary angles

never [62]1 year ago

5 0

Answer

You did not give the whole picture, soory

Step-by-step explanation:

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vladimir1956 [14]

Answer: 3.00

Step-by-step explanation:

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1 year ago

Solve for x 4x+6-7+9=18

Alla [95]

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2 years ago

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The figure here shows triangle AOC inscribed in the region cut from the parabola y=x^2 by the line y=a^2. Find the limit of the

aleksandrvk [35]

Area of the parabolic region = Integral of [a^2 - x^2 ]dx | from - a to a =

(a^2)x - (x^3)/3 | from - a to a = (a^2)(a) - (a^3)/3 - (a^2)(-a) + (-a^3)/3 =

= 2a^3 - 2(a^3)/3 = [4/3](a^3)

Area of the triangle = [1/2]base*height = [1/2](2a)(a)^2 = a^3

ratio area of the triangle / area of the parabolic region = a^3 / {[4/3](a^3)} =

Limit of a^3 / {[4/3](a^3)} as a -> 0 = 1 /(4/3) = 4/3

3 0

2 years ago

If anyone could help me, it would be greatly appreciated!

Sav [38]

Answer: $x=65^{\circ}, y=65^{\circ}, z=65^{\circ}$

Step-by-step explanation:

All radii of a circle are congruent, so by the base angles theorem

z = y = (180-50)/2 = 65.

Also, angles x and y are inscribed in the same arc, and they are thus congruent, meaning x = 65.

7 0

1 year ago

Find the derivative of the function f(x) = (x3 - 2x + 1)(x – 3) using the product rule.

julsineya [31]

Answer:

Step-by-step explanation:

Hello, first, let's use the product rule.

Derivative of uv is u'v + u v', so it gives:

$f(x)=(x^3-2x+1)(x-3)=u(x) \cdot v(x)\\\\f'(x)=u'(x)v(x)+u(x)v'(x)\\\\ \text{ **** } u(x)=x^3-2x+1 \ \ \ so \ \ \ u'(x)=3x^2-2\\\\\text{ **** } v(x)=x-3 \ \ \ so \ \ \ v'(x)=1\\\\f'(x)=(3x^2-2)(x-3)+(x^3-2x+1)(1)\\\\f'(x)=3x^3-9x^2-2x+6 + x^3-2x+1\\\\\boxed{f'(x)=4x^3-9x^2-4x+7}$

Now, we distribute the expression of f(x) and find the derivative afterwards.

$f(x)=(x^3-2x+1)(x-3)\\\\=x^4-2x^2+x-3x^3+6x-4\\\\=x^4-3x^3-2x^2+7x-4 \ \ \ so\\ \\\boxed{f'(x)=4x^3-9x^2-4x+7}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

6 0

2 years ago