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Mashcka [7]
3 years ago
11

The formula for a trapezoid relates the area A, the two bases, a and b, and the height, h. Solve for a

Mathematics
1 answer:
den301095 [7]3 years ago
8 0

Answer:

a=\frac{2A}{h}-b

Step-by-step explanation:

we know that

The formula to calculate the area for a trapezoid is equal to

A=\frac{1}{2}(a+b)h

where

a and b are the parallel bases

h is the height of trapezoid

A is the area

<em>Solve for a</em>

That means ----> isolate the variable a

Multiply by 2 both sides to remove the fraction in the right side

2A=(a+b)h

Divide by h both sides

\frac{2A}{h}=a+b

Subtract b both sides

\frac{2A}{h}-b=a

Rewrite

a=\frac{2A}{h}-b

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160=s.t

Faster train:(2) 160=(s+10) .(t-.5)From (1)(1) t=160/s
use the relations given in following screenshot and then use quadratic equation.

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A triangle has sides with lengths of 14 centimeters, 36 centimeters, and 39 centimeters. Is it a right triangle?
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4 years ago
Find all solutions to the equation in the interval [0,2pi). Enter the solutions in increasing order. Cos 2x = cos x
goblinko [34]

Step-by-step explanation:

\cos(2x)  =  \cos(x)

\cos {}^{2} (x)  -  \sin {}^{2} (x)  =  \cos(x)

\cos {}^{2} (x)   - (1 -  \cos {}^{2} (x) ) =  \cos(x)

2 \cos {}^{2} (x)  - 1 =  \cos(x)

2 \cos {}^{2} (x)  -  \cos(x)  - 1 = 0

2 \cos {}^{2} (x)   - 2 \cos(x)  +  \cos(x)  - 1 = 0

2 \cos(x) ( \cos(x)  - 1)  + 1( \cos(x)  + 1)

(2 \cos(x)   +  1)( \cos(x)  - 1) = 0

2 \cos(x)  + 1 = 0

2 \cos(x)  =  - 1

\cos(x)  =  -  \frac{1}{2}

x =   \frac{2\pi}{3} ,x =  \frac{4\pi}{3}

\cos(x)   - 1 = 0

\cos(x)  = 1

x = 0

So our answer are

0, \frac{2\pi}{3} , \frac{4\pi}{3}

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2 years ago
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