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lakkis [162]
3 years ago
12

I can’t find the surface area and volume of this cylinder, can someone please help?

Mathematics
2 answers:
tester [92]3 years ago
8 0

Answer:

Volume = 175.93km ^3

Surface Area =  753.97km ^2

Step-by-step explanation:

Cylinder surface area =A=2πrh+2πr2

Volume = V=πr2h

We plug in

Area = 2 x π x 8km x 7km =2π 56 = 351.85km ^2

= 351 858 377 m^2

+ 2πr2 = 2 x π x 64 = 402.12km^2

Surface Area = 2π 56 + 2π x 64

402.12 + 351.85 =753.97km ^2  = 753 97 8377 m^2

Volume = π x 8 x 8 x  7 = 448π = 1407.43km ^2

= 140.743m^3

stealth61 [152]3 years ago
7 0

Answer:

112 PIE CM CUBE

Step-by-step explanation:

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Help~~~~~~~~~~~~~~~~~~~~~~~~~
igor_vitrenko [27]

Answer:

28.25 square units

Step-by-step explanation:

A circumference of the circle is

C=2\pi r,

where r is the radius of the circle.

So,

18.84=2\pi r\\ \\r=\dfrac{18.84}{2\pi}=\dfrac{9.42}{\pi}\ cm

The area of the circle is

A=\pi r^2

Substitute the value of the radius:

A=\pi \cdot \left(\dfrac{9.42}{\pi}\right)^2=\dfrac{9.42^2}{\pi}\approx 28.25\ un^2

8 0
3 years ago
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How do you write 5.6 as a mixed number as a decimal?
Mama L [17]
5.6 in a fraction is 5 and 6/10, or 5 3/5. To get the mixed fraction, multiply the whole number (to the left) by the denominator (bottom of fraction), and add to the numerator (top of fraction). So 5 × 5 = 25 + 3 = 28. So it'd be 28/5.
4 0
2 years ago
Just before midnight, the first musher, William “Wild Bill” Shannon, left Nenana. It was -50ºF. When he took his first break, it
Lubov Fominskaja [6]

Answer:

-62°F

Step-by-step explanation:

it said it got colder so it got lower not higher

4 0
2 years ago
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Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
3 years ago
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Ierofanga [76]
Ask every fifth person that exits the school
(systematic random)
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