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ANEK [815]
3 years ago
8

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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Is this a question or is this spamming numbers ?

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Read 2 more answers
Find the surface area of the figure shown,<br> 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)

A=2\pi rh+2\pi r^2=2*\pi *4*9+2*\pi *4^2 = 326.73m^(cylinder)

326.73-96=230.73m^2 (\frac{3}{4} cylinder)

7 0
3 years ago
How long does it take to reach 2000$ with an investment of 500$ with a annual interest rate of 6% and doubles every 12 years?
In-s [12.5K]
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3 0
3 years ago
Which expression represents the volume of the prism?
GaryK [48]
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6 0
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
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