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3 years ago

What is the area of this trapezoid?

Mathematics

1 answer:

slamgirl [31]3 years ago

3 0

Area of rectangle in the middle = 12 * 9 = 108
Area of left triangle = .5 * 2 * 9 = 9
Area of right side triangle = .5 * 5 * 9 = 22.5

Total area = 108 + 9 + 22.5 = 139.5 square inches

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Using the given points, determine Δy.<br><br> (-3, -5) and (0, 10)

Jet001 [13]

Answer:

(-3,-5)

Step-by-step explanation:

because it as the information triangle y

3 0

2 years ago

A tortoise and a hare are competing in a 2000-meter race. The arrogant hare decides to let the tortoise have a 550-meter head st

andrew-mc [135]

Answer:

a) distance covered by hare d1 = 8t

b) distance covered by tortoise d2 = 5t + 550

c) ∆d = 550 - 3t

Step-by-step explanation:

Given;

Speed of hare u = 8m/s

Speed of tortoise v = 5 m/s

Initial distance of tortoise d0 = 550 m

a) using the equation of motion;

distance covered = speed × time + initial distance

d = vt + d0

For hare;

d0 = 0

Substituting the values;

d1 = 8t + 0

d1 = 8t

b)using the equation of motion;

distance covered = speed × time + initial distance

d2 = vt + d0

For tortoise;

d0 = 550m

Substituting the values;

d2 = 5t + 550

d2 = 5t + 550 m

c) the number of meters the tortoise is ahead of the hare.

∆d = distance covered by tortoise - distance covered by hare

∆d = d2 - d1

Substituting the values;

∆d = (5t + 550) - 8t

∆d = 550 - 3t

6 0

2 years ago

Mike decided to sell all of his old books he gathered up 9 books to sell he sold 2 books in a yard sale how many books does mike

Arlecino [84]

Answer: He has 7 books.

Step-by-step explanation:

9 - 2 = 7

8 0

2 years ago

Read 2 more answers

In a field, dry fodder for the cattle is heaped in a

ipn [44]

Answer:

volume of fodder=1/3pie R^2h= $\frac{1}{3} *\frac{22}{7} * (3.6)^2 *2.1=28.51m^3$

l^2=r^2+h^2

l^2=(2.1)^2+(3.6)^2

l^2=4.441+12.96=17.37

l= $\sqrt{17.37}$ =4.17m

minimum area of polythene to cover fodder=pieRl= $\frac{22}{7} *3.6*4.17=47.18m^2$

the volume of fodder is $28.51m^3$ and minimum area of polythene to cover fodder in the rainy seasonv47.18m^3

3 0

2 years ago

A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

4 0

2 years ago