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zhannawk [14.2K]

3 years ago

12

PLSSS HELP ME i need helppp

1 answer:

kozerog [31]3 years ago

3 0

Answer: 12.56

Explanation: 4x3.14

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What is the measure in radians for the central angle of a circle whose radius is 8 cm and intercepted arc length is 7.2 cm? Ente

natulia [17]

The measure in radians for the central angle of the circle is; 0.9 radians.

<h3>What is the angle measure in radians of the central angle?</h3>

Since, the length of the arc is given as 7.2cm and it's radius is 8cm.

It follows that the angle measure of the central angle can be evaluated as follows;

7.2 = (A/6.28) × 2× 3.14 × 8

7.2 = 8A

A = 7.2/8

A = 0.9 radians.

Read more on radian measure;

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8 0

2 years ago

Adult tickets to Space City amusement park cost x dollars. Children's tickets cost y dollars. The Garcia family bought 3 adult a

KengaRu [80]

let the number of adult tickets be x and the number of children tickets be y

3x + y = 164...equ(1)

2x + 3y = 174....equ(2)

multiplying equation 1 by 3

9x + 3y = 492

subtracting equation 2 from 1

7x = 318

x = 45.43 dollars

substituting the value of x into the equation

3(45.428) + y = 164

y = 164 - 3(45.428)

∴y = 27.71 dollars

8 0

3 years ago

Use the area of a rectangle and the Distributive<br> Property to find each product.<br> 9 X 34

OLga [1]

Answer:

Step-by-step explanation:

Write 34 in tens and units.

34 = 30 + 4

9(30 + 4)

9 * 30 = 270

9* 4 = 36

270 + 36 = 306

After awhile if your practice enough, you can do them in your head.

5 0

3 years ago

Adam is carrying books to the classroom for his teacher. Each books weighs 3.85 pounds. If he carries 4 books, how many pounds i

Elina [12.6K]

Adam is carrying 15.4 pounds because 3.85*4=15.4

5 0

4 years ago

Read 2 more answers

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

8 0

3 years ago