Multiply 4/9 x 1 & 4/5 x 2/5 A. 1 & 32/45 B. 7/16 C. 8/25 D. 1 & 10/19
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The product of the fraction is 8/25. Option C is correct
Multiplication of fraction
To multiply these fractions, you can follow these steps:
Multiply the numerators (top numbers) together.
Multiply the denominators (bottom numbers) together.
Here's how it's done:
(4/9) x (1 & 4/5) x (2/5)
Step 1: Multiply the numerators:
(4/9) x (9/5) x (2/5)
Step 2: Multiply the denominators:
(4/9) x (9/5) x (2/5) = (4 x 9 x 2)/(9 x 5 x 5)
Now, calculate the result:
(4 x 9 x 2)/(9 x 5 x 5) = 72/225
To simplify the fraction, find the greatest common divisor (GCD) of 72 and 225, which is 9:
(72 ÷ 9)/(225 ÷ 9) = 8/25
Hence the product of the fraction is 8/25
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Answers
A tessellation is created when a shape is repeated over and over again covering a plane without leaving any gaps or overlaps. Tessellation is also known as tiling. Triangles, squares and hexagon are perfect examples of figures that can create tessellation.
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I hope it helps, Regards.
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Laʀc = (2πR) • (θ ⁄ 2π) = R • θ ... where θ = central angle (of sector)
Ps = R + R + (R • θ) ... perimeter of sector = Ps
Ps = R • (2 + θ) ... Ps = 100 ft
100 = R • (2 + θ)
θ = (100 ⁄ R) – 2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~...
Similarly, the sector area is just a fraction of the circle area:
As = (πR²) • (θ ⁄ 2π) ... area of sector = As
As = R² • θ ⁄ 2 ... substitute for θ
As = R² • [(100 ⁄ R) – 2 ] ⁄ 2
As = 50R – R² ... differentiate
As' = 50 – 2R ... set to zero
0 = 50 – 2R
R = 25 ft ... optimum radius
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~...
As = 50R – R² ... evaluate at: R = 25 ft
As = 50(25) – (25)²
As = 625 ft² ... maximum area
Note ... at any other R_value, the sector area is less.
" θ " can be determined using: θ = (100 ⁄ R) – 2
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Final answer:
For a circular sector with a fixed perimeter of 100 ft, the values of radius (r) and arc length (s) that will maximize the sector's area are r=25 and s=50.
Explanation:
The perimeter of a circular sector is composed by the length of the arc (s) plus twice the radius (r). If this sum is fixed at 100 ft, then the length of the arc s is equal to 100 - 2r. The area A of a circular sector can be defined as A = 0.5 * r * s.
Substituting the expression for s into the area formula obtains A = 0.5 * r * (100-2r). Simplifying results in A = 50r - r^2 which is a downward opening parabola.
The maximum value of a parabola occurs at the vertex. For a parabola in the form y=ax^2 + bx + c, the x-coordinate of the vertex is -b/(2a). In this case, a=-1 and b=50, hence r=-50/2*(-1) = 25. Substituting r=25 back into the formula for s obtains s = 100-2*25 = 50. Therefore, the values for r and s that will give the circular sector the greatest area are r=25 and s=50.
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B. 48 sq meters
C. 24 sq meters
D. 7 sq meters
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191 times 21=191 times (20+1) correct?
therefor
use distributive property
remember that it is a(b+c)=ab+ac
therefor
191 times 21=191(20+1)=191 times 20+191 times 1=191 times 2 times 10+191=382 times 10+191=3820+191=4011
answer is 4011
×21
------
191
+382
---------
4011
answer= 4011
hope it helps
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Hi Sabra
2 3/4+ 3 1/5
= 5 19/20
I hope that's help ! Your answer for number 16 is incorrect
Sorry I don't know the answer for number 17 but I hope that's help.