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tensa zangetsu [6.8K]
3 years ago
14

NEED ANSWERS ASAP!! Help Pls I’d really appreciate it :)

Mathematics
2 answers:
alexandr1967 [171]3 years ago
7 0
Question 23:

The area of the shaded portion of the circle = area of the arc - area of the triangle
Area of arc = \frac{72}{360} * \pi r^2 =  \frac{72}{360}* \pi * 5^2 =  \frac{72*25}{360} \pi = 5 \pi
Area of triangle = 0.5 * base * height = 0.5 (2.9 + 2.9) * 4 = 11.6
Area of the shaded portion = (5π - 11.6) ft²
So, the correct answer is option A
=================================================
Question 24:
As shown in the problem the two congruent shaded figures have the following properties:
1. Each figure has four sides.
2. For each figure, each two opposite sides are parallel.

So, The correct statements are D and E
D. The figures are parallelograms because their opposite sides are parallel.
E. The figures are quadrilaterals because each has four sides.
=================================================
 Question 25:
The length of the line segment GH will be calculated as following:
Δ CEG is aright triangle at E
So the hypotenuse is CG = √(CE² + EG²) 
∴ CG = √(9² + 12²) = √225 = 15  →→ (1)
For the circle C
CH = CE = radius of the circle = 9 →→(2)
From (1) and (2)
∴ GH = CG - CH = 15 - 9 = 6

So, the correct answer is option D. 6 units
=================================================
 Question 26:
For the right cylinder
let the height of the cylinder = h
and let the radius of the base = r
so, the volume = π r² h

Given that: the height of the cylinder is 3 times the radius of the base∴ h = 3 r
Given the volume = 24π
∴ volume = π r² h = π r² * ( 3r ) = 3π r³
3π r³ = 24 π   ⇒⇒⇒ divide both sides over 3π
∴ r³ = (24π)/(3π) = 8
∴ r = ∛8 = 2
∴ h = 3r = 3 * 2 = 6

So, the correct answer is option C. 6 units
nika2105 [10]3 years ago
4 0
Hello,
Please, see the detailed solution in the attached files.
Thanks.

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Yuri [45]

The dimensions of the box are 10 ft and 5 ft

The maximum volume is 500 ft³

Step-by-step explanation:

A rectangle box with

  • A square base and no top
  • It needs to be made using 300 ft² of material
  • It has greatest volume

Surface area of a box without top (SA) = perimeter of base × height + area of the base

Volume of a box (V) = base area × height

Assume that the length of the side of the square base is x and the height of the box is y

∵ It needs to be made using 300 ft² of material

∴ The surface area of the box is 300 ft²

∵ Its base is a square of side length x ft

∴ Perimeter of the base = 4 × x = 4 x

∴ Area of the base = x²

∵ The height of the box = y ft

∵ SA = perimeter of base × height + area of the base

∵ SA = (4x)(y) + x²

∴ SA = 4xy + x²

∵ SA of the box = 300 ft²

- Equate the two expressions of SA

∴ 4xy + x² = 300

Now let us find y in terms of x

- Subtract x² from both sides

∴ 4xy = 300 - x²

- Divide each term by 4x to find y

∴ y=\frac{75}{x}-\frac{1}{4}x

∵ V = area of the base × height

∴ V = x² × y = x²y

- Substitute y by the equation of it above

∴ V=x^{2}(\frac{75}{x}-\frac{1}{4}x)

∴ V=75x-\frac{1}{4}x^{3}

∵ The volume of the box is greatest

- That means differentiate V and equate it by 0

∵ \frac{dV}{dx}=75-\frac{3}{4}x^{2}

∵ \frac{dV}{dx}=0 ⇒ greatest volume

∴ 75-\frac{3}{4}x^{2}=0

- Subtract 75 from both sides

∴ -\frac{3}{4}x^{2}=-75

- Divide both sides by -\frac{3}{4}

∴ x² = 100

- Take √ for both sides

∴ x = 10

Substitute the value of x in the equation of y

∵ y=\frac{75}{10}-\frac{1}{4}(10)

∴ y = 5

The dimensions of the box are 10 ft and 5 ft

∵ V=75x-\frac{1}{4}x^{3}

∵ x = 10

∴ V=75(10)-\frac{1}{4}(10)^{3}

∴ V=750-\frac{1}{4}(1000)

∴ V = 750 - 250

∴ V = 500 ft³

The maximum volume is 500 ft³

Learn more:

You can learn more about the volume in brainly.com/question/6443737

#LearnwithBrainly

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