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rewona [7]
3 years ago
11

Find the area of the following trapezoids b1=8 cm, b2=15 cm, h=11 cm

Mathematics
1 answer:
iVinArrow [24]3 years ago
3 0

Answer:

126.5 cm2

Step-by-step explanation:

First, the equation to find the area of a trapezoid is A= 1/2(b1 + b2)h  , where b1 is the first length and b2 is the second length and h is the height. For these measurements, you would simply factor them into the equation-  A= 1/2(8 + 15)11. Then using pemdas, 8+15= 23... multiplied by 1/2 equalling 11.5. Then multiply 11.5 by 11 to get your answer of 126.5 then add the units.

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Which step is the same in the construction of a perpendicular line through a point on a line and the construction of a perpendic
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Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e
svet-max [94.6K]

ANSWER

y = 2x -4

EXPLANATION

Part a)

Eliminating the parameter:

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x = 5 +  ln(t)

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From the first equation we make t the subject to get;

x - 5 =  ln(t)

t =  {e}^{x - 5}

We put it into the second equation.

y =  { ({e}^{x - 5}) }^{2}  + 5

y =  { ({e}^{2(x - 5)}) }  + 5

We differentiate to get;

\frac{dy}{dx}  = 2 {e}^{2(x - 5)}

At x=5,

\frac{dy}{dx}  = 2 {e}^{2(5 - 5)}

\frac{dy}{dx}  = 2 {e}^{0}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

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Without eliminating the parameter,

\frac{dy}{dx}  =  \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }

\frac{dy}{dx}  =  \frac{ 2t}{  \frac{1}{t} }

\frac{dy}{dx}  =  2 {t}^{2}

At x=5,

5 = 5 +  ln(t)

ln(t)  = 0

t =  {e}^{0}  = 1

This implies that,

\frac{dy}{dx}  =  2 {(1)}^{2}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

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