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3 years ago

What is the area of the trapezoid shown below?

Mathematics

2 answers:

Olin [163]3 years ago

7 0

Answer:

78 sqr units

Step-by-step explanation:

Use the Pythagorean Theorem to help find the base opposite the two.

The base of the second parallel line is sqrt( (15^2) - (12^2) ) = sqrt(81) = 9

The second base = 9 + 2 = 11

Area = (b1 + b2)*h/2

h = 12

b1 = 2

b2 = 11

h = 12

Area = (11 + 2)*12/2

Area = 13 * 12/2

Area = 78 square units

Murljashka [212]3 years ago

3 0

The answer is seventy eight square units

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2 Points

masya89 [10]

The answer is 120. Because inscribed angle is Half of the arc measure. Double 60 and we get 120.

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3 years ago

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What is the volume of the prism? 2 cm 3cm 11 cm PLS HURRY

mafiozo [28]

Answer:

Step-by-step explanation:

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2 years ago

Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=

luda_lava [24]

Let $f(x) = x^3 - 2x - 2$ . Then differentiating, we get

$f'(x) = 3x^2 - 2$

We approximate $f(x)$ at $x_1=2$ with the tangent line,

$f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18$

The $x$ -intercept for this approximation will be our next approximation for the root,

$10x - 18 = 0 \implies x_2 = \dfrac95$

Repeat this process. Approximate $f(x)$ at $x_2 = \frac95$ .

$f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}$

Then

$\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}$

Once more. Approximate $f(x)$ at $x_3$ .

$f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}$

Then

$\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}$

Compare this to the actual root of $f(x)$ , which is approximately 1.769292354, matching up to the first 5 digits after the decimal place.

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1 year ago

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netineya [11]

Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

What is the area of the trapezoid shown below?​

What is the area of the trapezoid shown below?