<img src="https://tex.z-dn.net/?f=4%2C%5Cpi%20%2C8" id="TexFormula1" title="4,\pi ,8" alt="4,\pi ,8" align="absmiddle" class="la

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Alisiya [41]

2 years ago

tex-formula"> how do i order from least to greatest

Mathematics

2 answers:

Yakvenalex [24]2 years ago

8 0

Answer:

$\pi, 4, 8$

Step-by-step explanation:

Note that $\pi\approx3.14$ . As $3$ , $\pi$ comes before $4$ .

Clearly $4$ , so $4$ comes before $8$ .

The only possible ordering is $\boxed{\pi, 4, 8}$

ivolga24 [154]2 years ago

5 0

Answer:

$\pi \: 4 \: 8$

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(5x+3)(5x+3)gghhhhhhhh

Stella [2.4K]

Answer:

25x^2 + 30x + 9

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2 years ago

What is the exact volume of the cylinder? enter your answer, in terms of pi, in the box.

Dominik [7]

Answer:

Step-by-step explanation:

r = 8 in

h = 5 in

Volume of cylinder = πr²h

= π * 8 * 8* 5

= 320π in³

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3 years ago

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What is 28% of 14,350

xenn [34]

Percent means parts out of 100 so
28%=28/100 or 0.28
'of' can be translated as multiply
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3 years ago

Read 2 more answers

If the volume of a box is 2x3 + 4x2 − 30xwhich of the dimensions are possible with the given x-value?

Kipish [7]

The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is $2 x^{3} + 4 x^{2} -30x$ .

Step-by-step explanation:

The given is,

$2 x^{3} + 4 x^{2} -30x$ ................................(1)

Step:1

Check for option A,

x = 1, dimensions 8 by 9 by 1

From the equation (1),

Volume = $2 (1^{3}) + 4 (1^{2} )-30(1)$

$=2+4-30$ = -24...................(2)

From the dimensions,

Volume = ( 8 × 9 × 1 )

= 72............................................(3)

From equation (2) and (3)

-24 ≠ 72

So, X=1; dimensions 8 by 9 by 1 is not possible.

Check for option B,

x = 1, dimensions 2 by 5 by 3

From the equation (1),

Volume = $2 (1^{3}) + 4 (1^{2} )-30(1)$

$=2+4-30$ = -24...................(4)

From the dimensions,

Volume = ( 2 × 5 × 3 )

= 30.........................................(5)

From equation (4) and (5)

-24 ≠ 30

So, X=1; dimensions 2 by 5 by 3 is not possible.

Check for option C,

x = 4, dimensions 2 by 5 by 3

From the equation (1),

Volume = $2 (4^{3}) + 4 (4^{2} )-30(4)$

$=2(64)+4(16)-30(4)$

$= 128+64-120$

= 72.............................................(6)

From the dimensions,

Volume = ( 2 × 5 × 3 )

= 30............................................(7)

From equation (6) and (7)

72 ≠ 30

So, X=4; dimensions 2 by 5 by 3 is not possible.

Check for option C,

x = 4, dimensions 8 by 9 by 1

From the equation (1),

Volume = $2 (4^{3}) + 4 (4^{2} )-30(4)$

$=2(64)+4(16)-30(4)$

$= 128+64-120$

= 72............................................(8)

From the dimensions,

Volume = ( 8 × 9 × 1 )

= 72............................................(9)

From equation (8) and (9)

72 = 72

So, X=4; dimensions 8 by 9 by 3 is possible.

Result:

The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is $2 x^{3} + 4 x^{2} -30x$ .

4 0

3 years ago

Please help me thank ykj

OLEGan [10]

F(4)= -4-8=-12
F(-2)=-(-2)-8=2-8= -6
F(2)= -2-8=-10
F(0)=0-8=-8
F(-3)= -(-3)-8=3-8=5

7 0

2 years ago