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shutvik [7]
3 years ago
9

PLEASE HELP!

Mathematics
1 answer:
Mashcka [7]3 years ago
4 0

Answer:

these are notes i wrote in geometry I'm not sure if this will help. it's been a few months since i did this

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What is the area of the trapozoid?
UkoKoshka [18]

Answer:

8units²

Step-by-step explanation:

area of trapezoid=½(a+b)*h

a=2

b=6

h=2

area of trapezoid=½(2+6)*2

area of trapezoid=½(8)*2

area of trapezoid=4*2

=8units²

5 0
3 years ago
What is the probability of rolling a number greater than 2 with a 6 sided die, three times in a row?
Fed [463]
A number greater than 2 is: 3, 4, 5, 6
Roll 1: 4 numbers out of 6 numbers total = 4/6 = 2/3
Roll 2: (same as roll 1)
Roll 3: (same as roll 1)

Roll 1 AND Roll 2 AND Roll 3
 2/3      x      2/3       x     2/3    = 2³/3³ = 8/27

Answer: \frac{8}{27}
3 0
3 years ago
To divide 3/8 ÷ 1/4 using a model, which of the following are true? Select all that are correct.
alina1380 [7]

So your answer according to these hints I provided for you, should lead you to your answer, which should in other words be,

The 1st. The 2nd. And I would say...The 4th. The reason why I say not the 3rd one is because the 4 has nothing to do with the 8, the 3 does because it's divided by it.

If it's not useful, feel free to report for mistake in answer. -_-



5 0
3 years ago
A spinner with 9 equal sections is numbered 1 through 9. The probability of spinning a 1 or a 9 is 2/9 .
USPshnik [31]
The answer would be 7/9. If the spinner has 9 sections and you want to know the probability of NOT landing on a 1 or a 9, you would subtract 2 from 9 since there are 2 numbers and 9 total.

6 0
3 years ago
Read 2 more answers
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
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