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posledela
3 years ago
15

Math.....................

Mathematics
2 answers:
Viefleur [7K]3 years ago
3 0
(A) would be your answer

Kamila [148]3 years ago
3 0

the answer to your question would be A

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puteri [66]

Answer:

yes

Step-by-step explanation:

yes it is im pretty sure

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3 years ago
Need an explanation
aleksandrvk [35]

Answer: Choice B

(6\sqrt{5}+5)i

==========================================================

Work Shown:

\sqrt{-5}+\sqrt{-25}+\sqrt{-125}\\\\\sqrt{-1*5}+\sqrt{-1*25}+\sqrt{-1*25*5}\\\\\sqrt{-1}*\sqrt{5}+\sqrt{-1}*\sqrt{25}+\sqrt{-1}*\sqrt{25}*\sqrt{5}\\\\i*\sqrt{5}+i*5+i*5*\sqrt{5}\\\\i*\sqrt{5}+5i+5i*\sqrt{5}\\\\(i\sqrt{5}+5i*\sqrt{5})+5i\\\\(\sqrt{5}+5\sqrt{5})i+5i\\\\(6\sqrt{5})i+5i\\\\(6\sqrt{5}+5)i\\\\

This points us to answer choice B

7 0
3 years ago
Cho hình chóp S.ABC có đáy abc là hình vuông tại A. Có ab = a, ac = a^3 . Mặt bên SAB là tam giác đều nằm trong măth phẳng vuông
Ira Lisetskai [31]

Answer:

wergi;bwfvhisbohouwdfkwv

Step-by-step explanation:

3 0
3 years ago
HELP PLZ FIRST ANSWER WILL GET BRAINLIEST<br> 5 5/6 times 4 1/4
Nikolay [14]

Hello!

It would be 24.79 but if the answer were to be rounded, it would be 25

Hope this helps! :)

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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