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ahrayia [7]
3 years ago
13

Ellie’s bird feeder is shaped like a cone with a height of 24 in. and a radius of 3 in. Packages of bird seed are cylindrical an

d come in different sizes. Solve for the height of the package that has a radius of 6 in. and contains exactly enough bird seed to fill the feeder. Use 3.14 to approximate pi.
Mathematics
1 answer:
love history [14]3 years ago
3 0
The volume of seeds in the package must be equal to the volume of the feeder in order to fill it completely. The volume of the conic feeder is given by:
V(f) = (πr²h)/3
And volume of the cylindrical package is given by:
V(p) = πr²h
Equating the two and substituting values:
(π x 3² x 24)/3 = π x 6² x h
h = 2 inches
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The perimeter of a rectangle is 15x+17y. if the length is 7/2x+7y, then find the width of the rectangle
Alexxandr [17]

Answer:

A

Step-by-step explanation:

Length : \frac{7x}{2} + 7y

Perimeter: 15x + 17y

Perimeter = 2(Length + Width)

15x + 17y = 2( \frac{7x}{2} + 7y + Width)

15x + 17y = 7x + 14y + 2(Width)

15x + 17y -7x -14y = 2(Width)

8x + 3y = 2(Width)

Width = \frac{8x + 3y}{2}

Width = 4x + 3y/2

3 0
3 years ago
5w+2=2w+5<br>have a good day!​
sveta [45]

Answer:

w=1

Step-by-step explanation:

5w+2=2w+5

3w+2=5

3w=3

w=1

4 0
3 years ago
Read 2 more answers
Asap rervfsdewrfgtsfdas
Sergeu [11.5K]

Step-by-step explanation:

QR = ST

⅔x = 0.4

times 3

2x = 1.2

x = 1.2 /2

x = 0.6

7 0
3 years ago
Read 2 more answers
The circumference of the base of a cone is 24 inches. The slant height of the cone is 20 inches.
loris [4]

The surface area of a cone with circumference base of 24π inches and slant height of 20 inches is 384π inches² in terms of π.

<h3>Surface area of a cone</h3>

surface area = πr(r + l)

where

  • r = radius
  • l = slant height

Therefore,

l = 20 inches

24π = 2πr

r = 12

Therefore,

surface area  = π(12)(12 + 20)

surface area = 12π(12 + 20)

surface area = 12π × 32

surface area = 384π inches²

learn more on surface area here: brainly.com/question/2847956

6 0
2 years ago
Read 2 more answers
Give an example and explain why a polynomial can have fewer x-intercepts than its number of roots.
NISA [10]
Answer:
A polynomial can have fewer x-intercepts than its number of roots when a pair of complex conjugate roots exist.

Example:
Consider the 4-th degree polynomial
f(x) = x⁴ - x³ - x² - x - 2

According to the Remainder theorem
f(-1) = 1 + 1 - 1 + 1 - 2 = 0
Therefore (x + 1) is a factor.

f(2) = 16 - 8 - 4 - 2 - 2 = 0
Therefore (x - 2) is a factor.

(x+1)*(x-2) = x² - 2x + x - 2 = x² - x - 2
To find the remaining factor, perform long division.
                            x²      + 1
          -------------------------------
x²-x-2 | x⁴  - x³  -  x²  - x  - 2
             x⁴  - x³ - 2x²
            -----------------------------
                             x² - x - 2
                             x² - x - 2
Therefore
f(x) = (x+1)(x-2)(x²+1)

Notice that (x² + 1) has no real factors.
However,
x² + 1 = (x + i)(x - i),
so it has a pair of conjugate zeros +i and -i.

A graph of the function confirms that there are only two real zeros (shown in red color).

7 0
3 years ago
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