Area of a trapezoid: 9 m by 15m by 20m

Mathematics

2 answers:

kvv77 [185]2 years ago

5 0

A=1/2 9+20×15
1/2. 29×15

217.5

exis [7]2 years ago

3 0

Equation:
$A = \frac{a + b}{2}h$

a (base) = 9
b (base) = 20
h (height) = 15

$A = \frac{9 + 20}{2}15$

$A = \frac{29}{2}15$

$A = 14.5 * 15$

$A = 217.5$

The area of the trapezoid is 217.5 m.

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What is the sum (Two-fifths x + StartFraction 5 over 8 EndFraction) + (one-fifth x minus one-fourth)?

liubo4ka [24]

The sum of ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) is $\frac{3}{5}$ x + $\frac{3}{8}$

Step-by-step explanation:

To add or subtract 2 fractions, they must have same denominators, if they not do that

Find the Lowest common multiple of the two denominators (LCM)
Replace each denominator by it
Divide The LCM by each denominator and multiply the numerator of each fraction by its quotient

∵ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ )

- Let us start with adding the like terms

∴ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) = (

∵ ( $\frac{2}{5}$ x + $\frac{1}{5}$ x ) have same denominators, then we can add them

∴ ( $\frac{2}{5}$ x + $\frac{1}{5}$ x ) = $\frac{3}{5}$ x

∵ ( $\frac{5}{8}$ - $\frac{1}{4}$ ) do not have the same denominators, then we must find

LCM of 8 and 4

- The LCM of 8 and 4 is 8 because 8 is the first common multiple

of 8 and 4

∵ LCM of 8 and 4 is 8

- Divide 8 by the denominator 4

∵ 8 ÷ 4 = 2