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Setler [38]
3 years ago
5

Find the circumference and area of a circle. Using 3.41 as the approximate value for X.

Mathematics
1 answer:
hichkok12 [17]3 years ago
3 0

Step-by-step explanation:

2*22/7*3.41=21.43

22/7*3.41*3.41=36.5454

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Justice hopes to build a fence around his garden he measures and determined that he needs 24 feet of fencing when he arrives at
Sunny_sXe [5.5K]

Answer:

It  will cost  $72 Justice to fence in his garden.

Step-by-step explanation:

Total length of the fencing around the garden =  24 feet

Cost of fencing per inch = $0.25

Now, as we know 1 foot  =  12 inches

⇒  24 feet =  24 x 12 inch  = 288 inches

⇒ 24 feet  =  288 inches

Hence, the total length of  fencing   =  288 inches.

Now, Total cost of Fencing

=Total length of fencing x Cost of fencing par inch

=  288 inches x ($0.25)

=  $ 72

or, the total cost of fencing = $72

Hence, it  will cost  $72 Justice to fence in his garden.

7 0
3 years ago
What is the volume of this cone?
olchik [2.2K]

Answer:

The volume of the cone is approximately 453.0 cm³

Step-by-step explanation:

The volume of a cone is one third that of a cylinder with the same height and radius.  That gives us  1/3 πr²h, where r is radius and h is height.

However, we are not given the height of the cone, but the side length.  We can work out the height using the Pythagorean theorem, as we have a right triangle with the height, base radius, and length.  You may recall that the Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of it's other two sides:

a^2 = b^2 + c^2

So we can find the height of the cone with that:

10.8^2 = 7.4^2 + h^2\\h^2 = 10.8^2 - 7.4^2\\h^2 = 116.64 - 54.76\\h^2 = 61.88\\h = \sqrt{61.88}\\h \approx  7.9

Now that we have the cone's height, we can solve for its volume:

v = \frac{1}{3} \pi r^2 h\\v = \frac{1}{3} \pi \times 7.4^2  \times 7.9\\v = \frac{1}{3} \pi \times 54.76 \times 7.9\\v = \frac{1}{3} \pi \times 432.6\\v = \pi \times 144.2\\v \approx 453.0 cm^3

5 0
3 years ago
Read 2 more answers
I wasn't in school today can someone explain how to do this
Pachacha [2.7K]
You have to multiply 7 3/4 to 9.

First, turn 7 3/4 to an improper fraction which is 31/4.
Next, you multiply 31/4 to 9/1(same thing as 9) across from each other.
Then, you get the answer 279/4(31 x 9 = 279 ;4 x 1 = 4; 279/4).
Last divide 279 to 4 and get 67.75( .75 = 3/4 ; 69 3/4).
 
7 0
4 years ago
"how many different rectangles" can you draw that have a perimeter of 12, 14
amm1812
Perimeter of rectangle = length + length + width + width

To find the combinations, think of two numbers that each multiplied by 2 and added up to give 12 or 14

Rectangle with perimeter 12

Say we take length = 2 and width = 3
Multiply the length by 2 = 2 × 2 = 4
Multiply the width by 3 = 2 × 3 = 6
Then add the answers = 4 + 6 = 10
This doesn't give us perimeter of 12 so we can't have the combination of length = 2 and width = 3

Take length = 4 and width = 2
Perimeter = 4+4+2+2 = 12
This is the first combination we can have

Take length = 5 and width = 1
Perimeter = 5+5+1+1 = 12
This is the second combination we can have

The question doesn't specify whether or not we are limited to use only integers, but if it is, we can only have two combinations of length and width that give perimeter of 12

length = 4 and width = 2
length = 5 and width = 1
--------------------------------------------------------------------------------------------------------------

Rectangle with perimeter of 14

Length = 4 and width = 3
Perimeter = 4+4+3+3 = 14

Length = 5 and width = 2
Perimeter = 5+5+2+2 = 14

Length = 6 and width = 1
Perimeter = 6+6+1+1 = 14

We can have 3 different combinations of length and width





3 0
3 years ago
the telephone company purchased 95 yards of wire for $1235 in september. they plan on making a purchase of 285 yards in October.
Luden [163]
Ratio and propoertion

cost/amount is constant
1235/95=x/285
cross multiply or time both sides by (95 times 283)
351975=95x
divide oth sies by 95
3705=x

the cost is $3705
7 0
3 years ago
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