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Kitty [74]
3 years ago
15

What's the product of 7/8 and 9/12​

Mathematics
2 answers:
kompoz [17]3 years ago
6 0

Answer:

21/32

Step-by-step explanation:

Goshia [24]3 years ago
3 0

Answer: 21/32 or 7/6

Step-by-step explanation:

depend on if you want the inverse operation

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A sector with a radius of 6 cm has a central angle measure of 4pi/9 (in radians). What is the area of the sector?
Usimov [2.4K]

Answer:

The area of the sector is 25.1327 cm²

Step-by-step explanation:

The area of a sector of circunference delimited by an angle in radians can be calculated by using the formula below:

area = (angle*r²)/2

area = [(4*pi/9)*(6)²]/2

area = [(4*pi/9)*36]/2

area = (144*pi/9)/2

area =  144*pi/18

area = 8*pi

area = 25.1327 cm²

The area of the sector is 25.1327 cm²

8 0
3 years ago
The Jansen family drove a total of 240 miles on their road trip. This distance is 5 times the distance they drove on the first d
kompoz [17]

Answer:

48 miles

Step-by-step explanation:

240/5 = 48

8 0
3 years ago
I need help with this question please!
katrin [286]
Need help with what, i cant draw the scale for you lol. 
5 0
3 years ago
HELP PLEASE
Marina CMI [18]

Answer:

1/20

Step-by-step explanation:

probability of numbers =1/20

  there are 20 numbers in the deck so each card represents 1/20.

7 0
3 years ago
Evaluate the integral by reversing the order of integration. Integral from 0 to 1 integral from √x to 1 of 5/(y³+1) dydx
ELEN [110]

Answer:

⁵/₃ ln 2

Step-by-step explanation:

∫₀¹ ∫√ₓ¹ 5 / (y³ + 1) dy dx

We want to change the order of integration.  To do this, we start by graphing the region contained by the four limits.

x = 0, x = 1

y = √x, y = 1

(Notice that y = √x is the same as x = y²).

Once we've graphed the region, we need to write the domain of x in terms of y.  In this case, 0 < x < y².

Then, we find the range: 0 < y < 1.

Now we can rewrite the integral:

∫₀¹ ∫₀ʸ² 5 / (y³ + 1) dx dy

Notice that the integrand itself, 5 / (y³ + 1), does not change.  Only the limits have changed.

Solve the integral.

∫₀¹ [ 5 / (y³ + 1) x |₀ʸ² ] dy

∫₀¹ [ 5y² / (y³ + 1) ] dy

⁵/₃ ∫₀¹ [ 3y² / (y³ + 1) ] dy

(⁵/₃ ln|y³ + 1|) |₀¹

⁵/₃ ln|1³ + 1| − ⁵/₃ ln|0³ + 1|

⁵/₃ ln 2

7 0
3 years ago
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