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Sergeu [11.5K]
3 years ago
10

What is 12/24, 3/30, 10/8, 34/20, and 14/8 in lowest term and in percent form

Mathematics
1 answer:
kirza4 [7]3 years ago
3 0
To find a fraction's lowest term, just simplify them and divide to find the decimal form.
12/24 = 1/2, 0.5
3/30 = 1/10, 0.1
10/8 = 5/4, 1.25
34/20 = 17/10, 1.7
14/8 = 7/4, 1.75

Finished!
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What is the simplified expression of (2 + 4) + (8 - 2) =​
Aleonysh [2.5K]

Answer:

12

Step-by-step explanation:

2+4=6

8-2=6

6+6=12

8 0
3 years ago
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The perimeter of a rectangle is 64 units. Can the length x of the rectangle can be 20 units when its width y is 11 units?
Oksi-84 [34.3K]
B. No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32

Perimeter = 2(l + w)
Perimeter = 64

Assuming: P = 64; w = 11 ; l = ?
64 = 2(l + 11)
64/2 = l + 11
32 - 11 = l
length = 21.

7 0
3 years ago
PLEASE HELP!! I am very confused. WILL GIVE BRAINLY!!!!
erastovalidia [21]

Answer:

x=80, y=15

Step-by-step explanation:

The angle on the opposite side of 100 will be equal to x

We know it is 180-100 so x=80

We know 6y=90

therfore y = 90/6 which equals 15

so y=15 and x=80

8 0
3 years ago
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I REALLY NEED HELP PLEASE HELP ME :(
o-na [289]

Answer:

I may be wrong but I think 8 is your answer.

Step-by-step explanation:

(-1)^(3/7)  x 128^(3/7)

-1 x 128^3/7

128^(3/7) = 8

= 8

6 0
3 years ago
Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.
rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed  as:

y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}

Thus, the area of the surface is as follows:

\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA

= \iint_D \sqrt{x^2 +y^2 +1 } \ dA

= \int^{2 \pi}_{0} \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr \ d \theta

=2 \pi \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr

= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0

= 2 \pi \times \dfrac{1}{3}  \Bigg [ (37)^{3/2} - 1 \Bigg]

= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]

3 0
2 years ago
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