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azamat
3 years ago
8

Suppose that a cylinder has a radius of r units, and that the height of the cylinder is also r units. The lateral area of the cy

linder is 338 square units. Find the value ofr.
Mathematics
1 answer:
kari74 [83]3 years ago
4 0

Answer:

r = 7.33 units

Step-by-step explanation:

Given that,

A cylinder has a radius of r units, and that the height of the cylinder is also r units.

The lateral area of the cylinder is 338 square units.

We need to find the value of r.

The formula for the lateral area of the cylinder is given by :

A=2\pi rh

Put all the values,

2\pi r^2=338\\\\r=\sqrt{\dfrac{338}{2\pi}} \\\\r=7.33\ unit

So, the value of r is equal to 7.33 units.

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What is the area of a triangle with side lengths 13-14-15?
Novosadov [1.4K]

84. Area of a triangle is (base x height) / 2

3 0
3 years ago
Solve the equation using the order of operations.<br><br> 5+{2×[(5−1)+6]}÷4
creativ13 [48]

Answer:

The order of operations is PEMDAS

First step is to do everything is parenthesis or brackets

5+{2*[(5-1)+6]}/4

{2*[(5-1)+6]}

First you would go 5-1=4, because that is the first equation that is in parenthesis by itself.

{2*[4+6]}

Next you would go 4+6=10, because that is your next smallest bracket

{2*10}

Last you would go 2*10=20

Your equation now looks like this

5+{20}/4

In PEMDAS your next step is exponents, but we don't have any so we go on to the next one which is multiply/division

5+5

You would go 20/4=5

Last step is to add/subtract

5+5=10

Your final answer would be 10

Hope this helps ;)

Step-by-step explanation:

5 0
3 years ago
I need help with this problem<br> 30 points up for grabs if you answer.
Rama09 [41]

Answer:

Step-by-step explanation:

13 cm

7 0
3 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
3 years ago
Dharmesh has a square garden with a perimeter of 132 feet. Is the area of the garden greater than 1,000 square feet?
Delvig [45]

Answer: Yes. The area is 1089 square feet.

Step-by-step explanation:

For a square, all the sides are equal: L = W

Hence, the perimeter of a square = 4 × L

Perimeter = 4L

132 = 4L

132 ÷ 4 = L

33 = L = W

Area of a square with 33 feet sides = L × L = 33 × 33 = 1089

1089 > 1000

3 0
4 years ago
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