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3 years ago

5

A cylinder has a height of 4 meters and a diameter that is 12 meters. Using 3.14 for pi, which of the following can be used to c

alculate the volume of the cylinder?
Group of answer choices

1 answer:

Studentka2010 [4]3 years ago

6 0

Answer:

Step-by-step explanation:

h = 4 m

d = 12 m

r = 12/2 = 6 m

Volume of cylinder = πr²h

= 3.14 * 6 * 6 * 4

= 452.16 m³

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Igoryamba

39. FALSE
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3 years ago

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What is the initial value of the equation shown? (1 point) y = −7x − 6 −13 −7 −6 −1

Liula [17]

Answer:

-6

Step-by-step explanation:

The equation is y = -7x - 6.

The initial value is found when x = 0.

y = -7(0) - 6

y = 0 - 6

y = -6

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2 years ago

Can someone please help me

astra-53 [7]

Answer:

square 20 has 44 green squares

square 21 has 45 green squares

Step-by-step explanation:

To solve the problem, we need to observe the cases, and determine/define a rule for each case (odd number of sides, or even number of sides).

For square one, we note that the centre square is shared by two diagonals, so we saved one square from the two diagonals.

The side length is 3 for square 1, 4 for square 2, and so on.

Let

n= square number (1, 2,3...)

L = side length (3,4,5...)

G1(n) = function that gives the number of green squares for square n, n=odd

G2(n) = function that gives the number of green squares for square n, n=even

side length, L=n+2 ................(1)

G1(n) = twice the side length less one, as discussed above

G1(n) = 2L-1 now substitute L=n+2

G1(n) = 2(n+2) -1 simplify

G1(n) = 2n + 3

Check:

for n=1, square 1 has 2*1+3 = 5 green squares ... checks

for n=3, square 3 has 2*3+3 = 9... checks

for n=5, square 5 has 2*5+3 = 13 ....checks

For even squares, it is even easier, because

G2(n) = 2L = 2(n+2)

check:

for n=2, square 2 has 2(2+2) = 8 green squares........checks

for n=4, square 4 has 2(4+2) = 12 green squares........checks.

Fincally, we apply our formula to n=20 and n=21

square 20 : G2(20) = 2(n+2) = 2(20+2) = 44 green quars

square 21 : G1(21) = 2n+3 = 2(21)+3 = 45 green squares

5 0

2 years ago

An adult ticket to a museum costs 3$ more than a children’s ticket. When 200 adult tickets and 100 children tickets are sold, th

antiseptic1488 [7]

The cost of children’s ticket is $ 5

<h3><u>Solution:</u></h3>

Let "c" be the cost of one children ticket

Let "a" be the cost of one adult ticket

Given that adult ticket to a museum costs 3$ more than a children’s ticket

<em>Cost of one adult ticket = 3 + cost of one children ticket</em>

a = 3 + c ------ eqn 1

<em><u>Given that 200 adult tickets and 100 children tickets are sold, the total revenue is $2100</u></em>

200 adult tickets x cost of one adult ticket + 100 children tickets x cost of one children ticket = 2100

$200 \times a + 100 \times c = 2100$

200a + 100c = 2100 ------ eqn 2

<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "c"</u></em>

Substitute eqn 1 in eqn 2

200(3 + c) + 100c = 2100

600 + 200c + 100c = 2100

600 + 300c = 2100

300c = 1500

<h3>c = 5</h3>

Thus the cost of children’s ticket is $ 5

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3 years ago

Simplify<br><br>-5|( 3 – 5•3^2)+2^3|•|4-2^3

aleksley [76]

Answer:

-232

Step-by-step explanation:

-5|+23|·|4-23|=-5|8|·|4-23|

-5|8|·|4-23|=-5|8|·|-4|

-5|8|·|-4|=-58·|-4|

-58·|-4|=-58·4

-58·4=-232

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3 years ago