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3 years ago

Find the area of each figure

Mathematics

1 answer:

Anon25 [30]3 years ago

7 0

Area of trapezoid = a + b/2 * h
a = 20, b = 9, h = 21/-16
20+9/2 * (21-16) = 29/2 * 5 = 72.5
The area of the trapezoid = 72.5 in^2

Area of the rectangle = l * w
l = 16, w = 20
16 * 20 = 320
The area of the rectangle: 320 in^2

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Find the surface area of the figure shown, correct to 2 decimal places.

agasfer [191]

Answer:

$230.73m^2$

Step-by-step explanation:

$A=2a^2+4ah=2*4^2+4*4*4=96m^2$ (square prism)

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3 years ago

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In-s [12.5K]

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3 years ago

Which expression represents the volume of the prism?

GaryK [48]

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3 years ago

Solve using the quadratic formula 3x^2+11x+5=0

storchak [24]

Answer:

$3x^2 + 11x + 5 = 0 : x = \frac{-11 + \sqrt{61} }{6} , x = \frac{-11 - \sqrt{61} }{6}$

Decimal:

$(x = -0.53162..., x = -3.13504...)$

Step-by-step explanation:

$3x^2+11x+5=0$

Solve with the quadratic formula:

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

For $a=3,\:b=11,\:c=5:\quad x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}$

$x=\frac{-11 + \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 + \sqrt{61} }{6}$

$x=\frac{-11 - \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 - \sqrt{61} }{6}$

The solutions to the quadratic equation are:

$x=\frac{-11+\sqrt{61}}{6},\:x=\frac{-11-\sqrt{61}}{6}$

Hope I helped. If so, may I get brainliest and a thanks?

Thank you, have a good day! =)

8 0

3 years ago