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loris [4]
2 years ago
8

What is the measure of RCP?

Mathematics
1 answer:
Bezzdna [24]2 years ago
6 0

Answer:

84

Step-by-step explanation:

15+15+27+27

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Find f(g(x)) and g(f(x)).<br> f(x)=x²-2 and g(x)=√x+1
Archy [21]

Answer:

1. x - 1

2. \sqrt{x^2-1}

Step-by-step explanation:

f(g(x)) = \sqrt{x+1}^2 - 2\\ = x + 1 - 2\\= x - 1

g(f(x)) =\sqrt{x^2-2+1}\\ = \sqrt{x^2-1}

6 0
2 years ago
Please help i need you
Natali [406]
The answer is A.0

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4 0
2 years ago
Pleaze hel0 me i dont know how to do this im giving a brainliest to the first person to answer​
Sliva [168]

Answer:

416mm^{3}

Step-by-step explanation:

Volume is found by Length x Width x Height

10.4 x 5 x 8      

416

7 0
3 years ago
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I will give brainliest answer
adell [148]

Answer: A=351.68 cm²

Step-by-step explanation:

To find the area of the shaded region, we would subtract the area of the circle by the area of the inner circle.

Area of Circle

A=\pi r^2

A=\pi (11)^2

A=121\pi

Area of inner (white) circle

A=\pi r^2

A=\pi (3)^2

A=9\pi

Now that we have the area to the circle and inner circle, we would subtract to find the area of the shaded region.

A=121\pi -9\pi

A=112\pi

A=351.68 cm^2

3 0
3 years ago
A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru
Viktor [21]

Answer:

1) S = 2\cdot w\cdot l - 8\cdot x^{2}, 2) The domain of S is 0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}. The range of S is 0 \leq S \leq 2\cdot w \cdot l, 3) S = 176\,in^{2}, 4) x \approx 4.528\,in, 5) S = 164.830\,in^{2}

Step-by-step explanation:

1) The function of the box is:

S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)

S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w

S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x

S = 2\cdot w\cdot l - 8\cdot x^{2}

2) The maximum cutout is:

2\cdot w \cdot l - 8\cdot x^{2} = 0

w\cdot l - 4\cdot x^{2} = 0

4\cdot x^{2} = w\cdot l

x = \frac{\sqrt{w\cdot l}}{2}

The domain of S is 0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}. The range of S is 0 \leq S \leq 2\cdot w \cdot l

3) The surface area when a 1'' x 1'' square is cut out is:

S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}

S = 176\,in^{2}

4) The size is found by solving the following second-order polynomial:

20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}

20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}

8\cdot x^{2} - 164\,in^{2} = 0

x \approx 4.528\,in

5) The equation of the box volume is:

V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x

V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x

V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}

V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}

V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}

The first derivative of the function is:

V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}

The critical points are determined by equalizing the derivative to zero:

12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0

x_{1} \approx 4.952\,in

x_{2}\approx 1.548\,in

The second derivative is found afterwards:

V'' = 24\cdot x - 78\,in

After evaluating each critical point, it follows that x_{1} is an absolute minimum and x_{2} is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

x \approx 1.548\,in

The surface area of the box is:

S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}

S = 164.830\,in^{2}

4 0
2 years ago
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